GIFT   OF 
heodore  Benedict  Lyinan 


0 

/sj"? 


WORKS  of  Professor  MANSFIELD  MERMAN 


PUBLISHED     BY 


JOHN  WILEY  &  SONS, 

43-45  East  Nineteenth  Street,  New  York. 

LONDON:    CHAPMAN   &   HALL,  LIMITED. 


Elements  of  Sanitary  Engineering.  For  the  use  of  Engineering 
Students  and  Municipal  Officers.  Octavo,  cloth.  Price  $2.00. 

Treatise  on  Hydraulics.  A  Text-book  for  Students  and  a  Manual 
for  Engineers.  Octavo,  cloth.  Price  $4.00. 

Elements  of  Precise  Surveying  and  Geodesy.     For   the   use   of 

Students  and  Civil  Engineers.     Octavo,  cloth.      Price  $2.50. 

Text-book  on  the  Method  of  Least  Squares.  Octavo,  cloth.  En- 
larged edition.  Price  $2.00. 

Mechanics   of   Materials  and  of   Beams,  Columns,   and   Shafts. 

Octavo,  cloth.     Enlarged  edition.     Price  $4.00. 

Strength  of  Materials.  An  Elementary  Text-book  for  Manual 
Training  Schools.  Duodecimo,  cloth.  Price  $1.00. 

Text-book  on  Roofs  and  Bridges.  By  Professors  MERRIMAN  and 
JACOBY.  In  Four  Parts.  Octavo,  cloth. 

PART     I.     Stresses  in  Simple  Trusses.     Price  $2.50. 
PART   II.     Graphic  Statics.     Price  $2.50. 
PART  II J,     Bridge  Design.     Price  $2.50. 
PART  IV.     Higher  Structures.     Price  $2. 50. 

Handbook  for  Surveyors.  By  Professors  MERRIMAN  and  BROOKS. 
Pocket-book  form,  morocco.  Price  $2.00. 

Higher  Mathematics.  A  Text-book  for  Classical  and  Engineering 
Colleges.  Edited  by  Professors  MERRIMAN  and  WOODWARD.  Octavo, 
cloth.  Price  $5.00. 


A   TEXT-BOOK 


ON   THE 


MECHANICS  OF  MATERIALS 


AND  OF 


BEAMS,  COLUMNS,  AND   SHAFTS. 


v  >;.**  ;.'  -..'  ;\  ;'>;  ; 
MANSFIELD   MERRIMAN, 

PROFESSOR   OF   CIVIL  ENGINEERING  IN   LEHIGH    UNIVERSITY. 


NINTH  EDITION,  REVISED. 
FOURTH   THOUSAND. 


NEW    YORK: 

JOHN    WILEY   &    SONS. 

LONDON:  CHAPMAN    &    HALL,   LIMITED. 

1903. 


COPYRIGHT,  1885,  1890,  1895, 
'MANSFIELD0'  M'ERRIMAN. 

G'<f  ft.    Vcr  *         «(>*crrre» 

I  TT        ^  T 


v  <^ 


\  c\ 


ROBERT  DRUMHONO,  KLECTROTYPBR  AND  PRINTER,  NBW  YORK 


PREFACE. 


The  following  pages  contain  an  elementary  course  of  study 
in  the  resistance  of  materials  and  the  mechanics  of  beams, 
columns  and  shafts,  designed  for  the  use  of  classes  in  technical 
schools  and  colleges.  It  should  be  preceded  by  a  good  train- 
ing in  mathematics  and  theoretical  mechanics,  and  be  followed 
by  a  special  study  of  the  properties  of  different  qualities  of 
materials,  and  by  detailed  exercises  in  construction  and  design. 

As  the  plan  of  the  book  is  to  deal  mainly  with  the  mechanics 
of  the  subject,  extended  tables  of  the  results  of  tests  on  different 
kinds  and  qualities  of  materials  are  not  given.  The  attempt, 
however,  has  been  made  to  state  average  values  of  the  quanti- 
ties which  express  the  strength  and  elasticity  of  what  may  be 
called  the  six  principal  materials.  On  account  of  the  great 
variation  of  these  values  in  different  grades  of  the  same  material 
the  wisdom  of  this  attempt  may  perhaps  be  questioned,  but 
the  experience  of  the  author  in  teaching  the  subject  during  the 
past  eleven  years  has  indicated  that  the  best  results  are  attained 
by  forming  at  first  a  definite  nucleus  in  the  mind  of  the  student, 
around  which  may  be  later  grouped  the  multitude  of  facts 
necessary  in  his  own  particular  department  of  study  and  work. 

As  the  aim  of  all  education  should  be  to  develop  the  powers 
of  the  mind  rather  than  impart  mere  information,  the  author 
has  endeavored  not  only  to  logically  set  forth  the  principles 
and  theory  of  the  subject,  but  to  so  arrange  the  matter  that 
students  will  be  encouraged  and  required  to  think  for  them- 
selves. The  problems  which  follow  each  article  will  be  found 


iii 


M23925 


IV  PREFACE. 

useful  for  this  purpose.  Without  the  solution  of  many  numer- 
ical problems  it  is  indeed  scarcely  possible  for  the  student  to 
become  well  grounded  in  the  theory.  The  attempt  has  been 
made  to  give  examples,  exercises,  and  problems  of  a  practical 
nature,  and  also  of  such  a  character  as  to  clearly  illustrate  the 
principles  of  the  theory  and  the  methods  of  investigation. 

MANSFIELD  MERRIMAN. 

SOUTH  BETHLEHEM*  PA.,  December,  1889. 


NOTE   TO   THE    NINTH    EDITION. 

The  sixth  edition  contained  double  the  matter  of  previous 
ones,  Chapters  VIII  to  XI  being  new.  In  the  seventh  edi- 
tion Arts.  6 1,  62,  117,  1 1 8,  149,  150,  and  151  were  rewritten. 
In  the  eighth  edition  medium  steel  was  substituted  for  wrought 
iron  in  the  discussion  of  all  rolled  I  beams.  In  this  edition 
all  known  errors  have  been  corrected  and  some  former  state- 
ments and  problems  replaced  by  better  ones.  New  matter 
on  eccentric  loads  on  columns  will  be  found  in  Art.  62,  on 
formulas  for  repeated  stresses  in  Art.  92,  on  spherical  and 
cylindrical  rollers  in  Arts.  106  and  107,  on  the  oscillations  of 
bars  and  beams  under  impact  in  Arts.  103  and  m£,  on  com- 
bined stresses  in  Art.  137,  and  on  the  International  Associa- 
tion for  Testing  Materials  in  Art.  152,  several  of  these  articles 
being  entirely  rewritten.  These  changes  and  other  minor 
ones  have  been  made  not  only  for  the  purpose  of  keeping  the 
book  abreast  with  modern  progress,  but  also  to  better  adapt 
it  to  the  use  of  students  and  engineers. 

M.  M. 


CONTENTS. 


CHAPTER  I. 

PAGES 

RESISTANCE   AND    ELASTICITY   OF   MATERIALS....     1-21 

Art.  i.  Average  Weights.  2.  Stresses  and  Deformations. 
3.  Experimental  Laws.  4.  Elastic  Limit  and  Coefficient  of 
Elasticity.  5.  Tension.  6.  Compression.  7.  Shear.  8.  Fac- 
tors of  Safety  and  Working  Stresses. 

CHAPTER  II. 

PIPES,  CYLINDERS,    AND   RIVETED   JOINTS 22-35 

Art.  9.  Water  and  Steam  Pipes.  10.  Thin  Cylinders  and 
Spheres,  u.  Thick  Cylinders.  12.  Investigation  of  Riveted 
Joints.  13.  Design  of  Riveted  Joints.  14.  Miscellaneous 
Exercises. 

CHAPTER  III. 

CANTILEVER   BEAMS   AND   SIMPLE   BEAMS 36-84 

Art.  15.  Definitions.  16.  Reactions  of  the  Supports.  17. 
The  Vertical  Shear.  18.  The  Bending  Moment.  19.  Inter- 
nal Stresses  and  External  Forces.  20.  Experimental  and 
Theoretical  Laws.  21.  The  Two  Fundamental  Formulas. 
22.  Center  of  Gravity  of  Cross-sections.  23.  Moment  of  Iner- 
tia of  Cross-sections.  24.  The  Maximum  Bending  Moment. 
25.  The  Investigation  of  Beams.  26.  Safe  Loads  for  Beams. 
27.  Designing  of  Beams.  28.  The  Modulus  of  Rupture.  29. 
Comparative  Strengths.  30.  Iron  and  Steel  I  Beams.  31. 
Iron  and  Steel  Deck  Beams.  32.  Cast-iron  Beams.  33.  Gen- 
eral Equation  of  the  Elastic  Curve.  34.  Deflection  of  Canti- 
lever Beams.  35.  Deflection  of  Simple  Beams.  36.  Compar- 
ative Deflection  and  Stiffness'.  37.  Relation  between  Deflec- 
tion and  Stress.  38.  Cantilever  Beams  of  Uniform  Strength. 
39.  Simple  Beams  of  Uniform  Strength. 


Vi  CONTENTS. 

CHAPTER  IV. 

PACKS 

RESTRAINED    BEAMS   AND   CONTINUOUS   BEAMS.     85-110 

Art.  40.  Beams  Overhanging  One  Support.  41.  Beams 
Fixed  at  One  End  and  Supported  at  the  Other.  42.  Beams 
Overhanging  Both  Supports.  43.  Beams  Fixed  at  Both 
Ends.  44.  Comparison  of  Restrained  and  Simple  Beams. 
45.  General  Principles  of  Continuity.  46.  Properties  of 
Continuous  Beams.  47.  The  Theorem  of  Three  Moments. 
48.  Continuous  Beams  with  Equal  Spans.  49.  Continuous 
Beams  with  Unequal  Spans.  50.  Remarks  on  the  Theory 
of  Flexure. 

CHAPTER  V. 

COLUMNS   OR   STRUTS 111-134 

Art.  51.  Cross-sections  of  Columns.  52.  General  Princi- 
ples. 53.  EULER'S  Formula.  54.  HODGKINSON'S  Formu- 
las. 55.  RANKINE'S  Formula.  56.  Radius  of  Gyration  of 
Columns.  57.  Investigation  of  Columns.  58.  Safe  Loads 
for  Columns.  59.  Designing  of  Columns.  60.  The  Straight- 
line  Formula.  61.  RITTER'S  Rational  Formula.  62.  Ec- 
centric Loads. 

CHAPTER  VI. 

TORSION    AND    SHAFTS I35-I43 

Art.  63.  The  Phenomena  of  Torsion  64.  The  Funda- 
mental Formula  for  Torsion.  65.  Polar  Moments  of  Iner- 
tia. 66.  The  Constants  of  Torsion.  67.  Shafts  for  Trans- 
mission of  Power.  68.  Round  Shafts.  69.  Hollow  Shafts. 
70.  Miscellaneous  Exercises. 

CHAPTER  VII. 

COMBINED   STRESSES 144-161 

Art.  71.  Combined  Tension  and  Compression.  72. 
Stresses  due  to  Temperature.  73.  Combined  Tension  and 
Flexure.  74.  Combined  Compression  and  Flexure.  75. 
Shear  Combined  with  Tension  or  Compression.  76.  Com- 


CONTENTS.  Vll 

PAGES 

bined  Flexure  and  Torsion.  77.  Combined  Compression 
and  Torsion.  78.  Horizontal  Shear  in  Beams.  79.  Maxi- 
mum Internal  Stresses  in  Beams. 

CHAPTER  VIII. 

THE   STRENGTH    OF   MATERIALS.. 162-196 

Art.  80.  Mean  Constants.  81.  Historical.  82.  Testing 
Machines.  83.  Timber.  84.  Brick.  85.  Cement  and  Mor- 
tar. 86.  Stone.  87.  Cast  Iron.  88.  Wrought  Iron.  89. 
Steel.  90.  Other  Materials.  91.  The  Fatigue  of  Materials. 
92.  Repeated  Stresses. 

CHAPTER  IX. 

THE    RESILIENCE  OF  MATERIALS 197-221 

Art.  93.  Sudden  Loads  and  Impact.  94.  The  Modulus  of 
Resilience.  95.  External  Work  and  Resilience.  96.  Elastic 
Resilience  of  Beams.  97.  Ultimate  Resilience.  98.  Early 
History  of  Resilience.  99.  Modern  Experiments. 

CHAPTER   X. 

TENSION   AND   COMPRESSION 222-240 

Art.  ico.  Elongation  under  own  Weight.  101.  Bar  of 
Uniform  Strength.  102.  Longitudinal  Impact.  103.  Oscil- 
lations after  Impact.  104.  Centrifugal  Stress.  105.  Shrink- 
age of  Hoops.  1 06.  Spherical  Rollers.  107.  Cylindrical 
Rollers.  108.  Eccentric  Loads. 

CHAPTER   XL 

FLEXURE   OF   BEAMS 241-271 

Art.  109.  The  Work  of  Flexure,  no.  Static  and  Sudden 
Deflections,  in.  Deflection  under  Impact.  m|.  Oscilla- 
tions after  Impact.  112.  Pressure  due  to  Impact.  113.  Cen- 
trifugal Stress.  114.  Live-load  Velocity.  115.  Work  of  the 
Vertical  Shear.  116.  Deflection  due  to  Shearing.  117.  Flex- 
ure and  Compression.  118.  Flexure  and  Tension. 


viii  CONTENTS. 

CHAPTER  XII. 

PAGES 

SHEAR  AND  TORSION ,272-287 

Art.  119.  Stresses  due  to  Shear.  120.  Resilience  under 
Shear.  121.  The  Coefficient  of  Elasticity.  122.  Resilience 
under  Torsion.  123.  Hollow  Shafts.  124.  Shaft  Couplings. 
125.  A  Crank  Pin  and  Shaft.  126.  A  Triple-crank  Pin. 

CHAPTER  XIII. 

APPARENT  STRESSES  AND  TRUE  STRESSES...  288-309 
Art.  127.  Mathematical  Theory  of  Elasticity.  128.  Lateral 
Deformation.  129.  True  Tensile  and  Compressive  Stresses. 
130.  Normal  and  Tangential  Stresses.  131.  Resultant 
Stresses.  132.  The  Ellipsoid  of  Stress.  133.  The  Three 
Principal  Stresses.  134.  A  Numerical  Case.  135.  Maxi- 
mum Shearing  Stresses.  136.  The  Ellipse  of  Stress.  137. 
Formulas  for  True  Combined  Stresses. 

CHAPTER  XIV. 

STRESSES  IN  GUNS. 310-327 

Art.  138.  LAMP'S  Formulas.  139.  A  Solid  Gun.  140.  A 
Compound  Cylinder.  141.  CLAVARINO'S  Formulas.  142. 
BIRNIE'S  Formulas.  143.  Hoop  Shrinkage.  144.  Hooped 
Guns. 

CHAPTER  XV. 

PLATES,  SPHERES,  AND  COLUMNS.. .  328-343 

Art.  145.  Circular  Plates.  146.  Elliptical  Plates.  147. 
Rectangular  Plates.  148.  Hollow  Spheres.  149.  Experi- 
ments on  Columns.  150.  EULER'S  Modified  Formula.  151. 
The  Deflection  of  Columns. 

APPENDIX 344-362 

International  Association  for  Testing  Materials.  The  Ve- 
locity of  Stress.  Advanced  Problems.  Answers  to  Prob- 
lems. Description  of  Tables.  Tables  of  Logarithms,  Squares, 
and  Circles.  Weights  of  Wrought-iron  Bars. 

INDEX , 363-368 


CONTENTS.  IX 

TABLES. 

PAGES 

Average  Weights  and  Specific  Gravities i 

Constants  for  Tension 9 

Tensile  Test  of  a  Wrought-iron  Bar 1 1 

Constants  for  Compression 14 

Constants  for  Shearing 15 

Factors  of  Safety 1 8 

Centers  of  Gravity  of  Cross-sections 52 

Moments  of  Inertia  of  Cross-sections 53 

Properties  of  I  Beams 65 

Properties  of  Deck  Beams 67 

Cantilever  Beams  and  Simple  Beams 79 

Cantilever,  Simple,  and  Restrained  Beams 95 

Continuous  Beams  with  Equal  Spans 104,  105 

Average  Constants  for  Columns 122 

Radii  of  Gyration  of  Cross-sections 1 23 

The  Straight-line  Column  Formula f . . . .  128 

Experiments  on  Rupture  of  Columns 130 

Weights  and  Coefficients  of  Expansion 163 

Elastic  Limits  and  Coefficients  of  Elasticity 163 

Ultimate  Tensile  and  Compressive  Strengths 164 

Ultimate  Shearing  Strength  and  Modulus  of  Rupture 164 

Strength  of  Timber    171 

Strength  of  Cement  and  Mortar 175 

Strength  of  Stone 177 

Strength  of  Wrought  Iron 183 

Strength  of  Steel 187 

Ultimate  Resilience  of  a  Steel  Bar 206 

Velocities  of  Stress 347 

Logarithms  of  Numbers 356 

Squares  of  Numbers 358 

Areas  of  Circles 360 

Weights  of  Wrought-iron  Bars 362 


Evolvi  varia  problemata.  In  scientiis  enim  ediscendis 
prosunt  exempla  magis  quam  praecepta.  Qua  de  causa  in  his 
fusius  expatiatus  sum.— NEWTON. 


Nous  avons  pour  but,  non  dc  donner  un  traitecomplet,  mais 
de  montrer,  par  dcs  exemples  simples  et  varies,  I'utilit6  et  1'im- 
portance  de  la  theorie  mathematique  de  1'elasticite. — LAM* 


MECHANICS  OF  MATERIALS. 


CHAPTER   I. 
THE    RESISTANCE   AND    ELASTICITY   OF   MATERIALS. 

ARTICLE  i.    AVERAGE  WEIGHTS. 

The  principal  materials  used  in  engineering  constructions 
are  timber,  brick,  stone,  cast  iron,  wrought  iron,  and  steeJ. 
The  following  table  gives  their  average  unit-weights  and  aver- 
age specific  gravities. 


Average  Weight. 

Material. 

Average 
Specific 
Gravity. 

Pounds  per 
Cubic  Foot. 

Kilos  per 
Cubic  Meter. 

Timber 

40 

600 

0.6          | 

Brick 

125 

2000 

2.0 

Stone 

TOO 

2  560 

2.6 

Cast  Iron 

450 

7  200 

7.2 

Wrought  Iron 

480 

7700 

7-7 

Steel 

49° 

7  800 

7.8 

These  weights,  being  mean  or  average  values,  should  be  care- 
fully memorized  by  the  student  as  a  basis  for  more  precise 
knowledge,  but  it  must  be  noted  that  they  are  subject  to  more 
or  less  variation  according  to  the  quality  of  the  material. 
Brick,  for  instance,  may  weigh  as  low  as  100,  or  as  high  as  150 
pounds  per  cubic  foot,  according  as  it  is  soft  or  hard  pressed* 


RESISTANCE   AND   ELASTICITY   OF   MATERIALS.      CH.  I. 

Unless  otherwise  stated  the  above  average  values  will  be  used 
in  the  examples  and  problems  of  this  book.  In  all  engineering 
reference  books  are  given  tables  showing  the  unit-weights  far 
different  qualities  of  the  above  six  principal  materials,  and  also 
for  copper,  lead,  glass,  cements,  and  other  materials  used  in 
construction. 

\  For\cpiji^*4t4ng  the  weights  of  bars,  beams,  and  pieces  of  uni- 
t  t  e  form<  cross-section,. J:he  following  approximate  simple  rules  will 
A I  ii'  bft^n,  be;  fauhd^CenVenient. 

A  wrought  iron  bar  one  square  inch  in  section  and  one 

yard  long  weighs  ten  pounds. 

Steel  is  about  two  per  cent  heavier  than  wrought  iron. 
Cast  iron  is  about  six  per  cent  lighter  than  wrought  iron. 
Stone  is  about  one-third  the  weight  of  wrought  iron. 
Brick  is  about  one-fourth  the  weight  of  wrought  iron. 
Timber  is  about  one-twelfth  the  weight  of  wrought  iron. 
For  example,  consider  a  bar  of  wrought  iron  i£  X  3  inches  and 

12  feet  long;   its  cross-section  is  4.5  square  inches,  hence  its 
weight    is   45  X  4  =  180  pounds.     A  steel    bar  of   the   same 
dimensions  will  weigh  180  +  0.02  X  180  =  about  184  pounds, 
and  a  cast  iron  bar  will  weigh  180  —  0.06  X  180  —  about  169 
pounds. 

By  reversing  the  above  rules  the  cross-sections  of  bars  are 
readily  computed  from  their  weights  per  yard.  Thus,  if  a  stick 
of  timber  15  feet  long  weigh  120  pounds,  its  weight  per  yard  is 
24  pounds,  and  its  cross-section  is  12  X  2.4  =  about  28.8  square 
inches. 

Problem  I.  How  many  square  inches  in  the  cross-section  of  a 
wrought  iron  railroad  rail  weighing  24  pounds  per  linear  foot  ? 
In  a  steel  rail?  In  a  wooden  beam? 

Prob.  2.  Find  the  weights  of  a  wooden  beam  6x8  inches  in 
section  and  13  feet  long,  of  a  steel  bar  one  inch  in  diameter  and 

13  feet  long,  and  of  a  common  brick  2X4  inches  and  8  inches 
long. 


ART.  2.  STRESSES   AND   DEFORMATIONS.  3 

ART.  2.    STRESSES  AND  DEFORMATIONS. 

A  'stress '  is  a  force  which  acts  in  the  interior  of  a  body  and 
resists  the  external  forces  which  tend  to  change  its  shape.  If 
a  weight  of  400  pounds  be  suspended  by  a  rope,  the  stress  in 
the  rope  is  400  pounds.  This  stress  is  accompanied  by  an 
elongation  of  the  rope,  which  increases  until  the  internal  molec- 
ular stresses  or  resistances  are  in  equilibrium  with  the  exterior 
weight.  Stresses  are  measured  in  pounds,  tons,  or  kilograms. 
A  'unit-stress'  is  the  amount  of  stress  on  a  unit  of  area;  this  is 
expressed  either  in  pounds  per  square  inch,  or  in  kilograms  per 
square  centimeter.  Thus,  if  a  rope  of  two  square  inches  cross- 
section  sustains  a  stress  of  400  pounds,  the  unit-stress  is  200 
pounds  per  square  inch,  for  the  total  stress  must  be  regarded 
as  distributed  over  the  two  square  inches  of  cross-section. 

A  '  deformation '  is  the  amount  of  change  of  shape  of  a 
body  caused  by  the  external  forces.  If  a  load  be  put  on  a 
column  its  length  is  shortened,  and  the  amount  of  shortening 
is  a  deformation.  So  in  the  case  of  the  rope,  the  amount  of 
elongation  is  a  deformation.  Deformations  are  generally  meas- 
ured in  inches,  or  centimeters. 

The  word  '  strain '  is  often  used  in  technical  literature  as 
synonymous  with  stress,  and  sometimes  it  is  also  used  to  desig- 
nate the  deformation,  or  change  of  shape.  On  account  of  this 
ambiguity  the  word  will  not  be  employed  in  this  book. 

Three  kinds  of  simple  stress  are  produced  by  forces  which 
tend  to  change  the  shape  of  a  body.  They  are, 

Tensile,  tending  to  pull  apart,  as  in  a  rope. 
Compressive,  tending  to  push  together,  as  in  a  column. 
Shearing,  tending  to  cut  across,  as  in  punching  a  plate. 

The  nouns  corresponding  to  these  three  adjectives  are  Tension, 
Compression,  and  Shear.     The  stresses  which  occur  in  beams, 


4  RESISTANCE  AND   ELASTICITY  OF   MATERIALS.      CH.  I. 

columns,  and  shafts  are  of  a  complex  character,  but  they  may 
always  be  resolved  into  the  three  kinds  of  simple  stress.  The 
first  effect  of  an  applied  force  is  to  cause  a  deformation. 
This  deformation  receives  a  special  name  according  to  the  kind 
of  stress  which  accompanies  it.  Thus, 

Tension  produces  an  elongation. 
Compression  produces  a  shortening. 
Shear  produces  a  detrusion. 

This  change  of  shape  is  resisted  by  the  stresses  between  the 
molecules  of  the  body,  and  as  soon  as  these  internal  resistances 
balance  the  exterior  forces  the  change  of  shape  ceases  and  the 
body  is  in  equilibrium.  But  if  the  external  forces  be  increased 
far  enough  the  molecular  resistances  are  finally  overcome  and 
the  body  breaks  or  ruptures. 

In  any  case  of  simple  stress  in  a  body  in  equilibrium  the 
total  internal  stresses  or  resistances  must  equal  the  external 
applied  force.  Thus,  in  the  above  instance  of  a  rope  from 
which  a  weight  of  400  pounds  is  suspended,  let  it  be  imagined 
to  be  cut  at  any  section ;  then  equilibrium  can  only  be  main- 
tained by  applying  at  that  section  an  upward  force  of  400 
pounds ;  hence  the  stresses  in  that  section  must  also  equal  400 
pounds.  In  general,  if  a  steady  force  P  produce  either  ten- 
sion, compression,  or  shear,  the  total  stress  produced  is  also  P+ 
for  if  not  equilibrium  does  not  obtain.  In  such  cases,  then,  the 
word  *  stress '  may  be  used  to  designate  the  external  force  as 
well  as  the  internal  resistances. 

Tension  and  Compression  are  similar  in  character  but  differ 
in  regard  to  direction.  A  tensile  stress  in  a  bar  occurs  when 
two  forces  of  equal  intensity  act  upon  its  ends,  each  in  a  direc- 
tion away  from  the  other.  In  compression  the  direction  of  the 
forces  is  reversed  and  each  acts  toward  the  bar.  Evidently  a 
simple  tensile  or  compressive  stress  in  a  bar  is  to  be  regarded 
as  evenly  distributed  over  the  area  of  its  cross-section,  so  that 


ART.  2.  STRESSES   AND   DEFORMATIONS.  5 

if  P  be  the  total  stress  in  pounds  and  A  the  area  of  the  cross- 

P 

section  in  inches,  the  unit-stress  is  -j  in  pounds  per  square  inch. 

A 

Shear  requires  the  action  of  two  forces  exerted  in  parallel 
planes  and  very  near  together,  like  the  forces  in  a  pair  of 
shears,  from  which  analogy  the  name  is  derived.  Here  also 
the  total  shearing  stress  P  is  to  be  regarded  as  distributed 

P 

uniformly  over  the  area  A,  so  that  the  unit-stress  is    -r-      And 

A 

conversely  if  ^S  represent  the  uniform  unit-stress  the  total  stress 
Pis  AS. 

In  any  case  of  simple  stress  acting  on  a  body  let  P  be  the 
total  stress,  A  the  area  over  which  it  is  uniformly  distributed, 
and  5  the  unit-stress.  Then, 

(i)  P=AS. 

Also  let  A  be  the  total  linear  deformation  produced  by  the 
stress,  /  the  length  of  the  bar,  and  s  the  deformation  per  unit 
of  length.  Then  this  deformation  is  to  be  regarded  as  uni- 
formly distributed  over  the  distance  /,  so  that  also, 

(i)'  \  =  Is. 

The  laws  implied  in  the  statement  of  these  two  formulas  are 
confirmed  by  experiment,  if  the  stress  be  not  too  great. 

Unit-stress  in  general  will  be  denoted  by  5,  whether  it  be 
tension,  compression,  or  shear.  S,  will  denote  tensile  unit- 
stress,.  Sc  compressive  unit-stress,  and  St  shearing  unit-stress, 
when  it  is  necessary  to  distinguish  between  them. 

Prob.  3.  A  wrought  iron  rod  ij  inches  in  diameter  breaks 
under  a  tension  of  67610  pounds.  Find  the  breaking  unit- 
stress. 

Prob.  4.  If  a  cast-iron  bar  i£  X  2  inches  in  size  breaks 
under  a  tension  of  60  ooo  pounds,  what  tension  will  break  a 
bar  if  X  if  inches  in  size? 


RESISTANCE   AND   ELASTICITY    OF   MATERIALS.      CH.  L 


ART.  3.     EXPERIMENTAL  LAWS. 

Numerous  tests  or  experiments  have  been  made  to  ascertain 
the  strength  of  materials  and  the  laws  that  govern  stresses  and 
deformations.  The  resistance  of  a  rope,  for  instance,  may  be 
investigated  by  suspending  it  from  one  end  and  applying 
weights  to  the  other.  As  the  weights  are  added  the  rope  will 
be  seen  to  stretch  or  elongate,  and  the  amount  of  this  deforma- 
tion may  be  measured.  When  the  load  is  made  great  enough 
the  rope  will  break,  and  thus  its  ultimate  tensile  stress  is- 
known.  For  stone,  iron,  or  steel,  special  machines,  known  as 
testing  machines,  have  been  constructed  by  which  the  effect  of 
different  stresses  on  different  qualities  and  forms  of  materials 
may  be  accurately  measured. 

All  experiments,  and  all  experience,  agree  in  establishing  the 
five  following  laws  for  cases  of  simple  tension  and  compression, 
which  may  be  regarded  as  the  fundamental  principles  of  the 
science  of  the  strength  of  materials. 

(A) — When  a  small  stress  is  caused  in  a  body  a  small  de- 
formation is  produced,  and  on  the  removal  of  the  stress 
the  body  springs  back  to  its  original  form.  For  small 
stresses,  then,  materials  may  be  regarded  as  perfectly 
elastic. 

(B] — Under  small  stresses  the  deformations  are  approxi- 
mately proportional  to  the  forces,  or  stresses,  which  pro- 
duce them,  and  also  approximately  proportional  to  the 
length  of  the  bar  or  body. 

(C) — When  the  stress  is  great  enough  a  deformation  is 
produced  which  is  partly  permanent,  that  is,  the  body 
does  not  spring  back  entirely  to  its  original  form  on  re- 
moval of  the  stress.  This  permanent  part  is  termed  a 
set.  In  such  cases  the  deformations  are  not  proportional 
to  the  stresses. 


ART.  4.    ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY.       / 

(D] — When  the  stress  is  greater  still  the  deformation 
rapidly  increases  and  the  body  finally  ruptures. 

(E) — A  sudden  stress,  or  shock,  is  more  injurious  than  a 
steady  stress  or  than  a  stress  gradually  applied. 

The  words  small  and  great,  used  in  stating  these  laws,  have,  as 
will  be  seen  later,  very  different  values  and  limits  for  different 
kinds  of  materials  and  stresses. 

The 'ultimate  strength '  of  a  material  under  tension,  com- 
pression, or  shear,  is  the  greatest  unit-stress  to  which  it  can  be 
subjected.  This  occurs  at  or  shortly  before  rupture,  and  its 
value  is  very  different  for  different  materials.  Thus  if  a  bar 
whose  cross-section  is  A  breaks  under  a  tensile  stress  P,  the 
ultimate  tensile  strength  of  the  material  is  P -r-  A. 

Prob.  5.  If  the  ultimate  strength  of  wrought  iron  is  55  ooo 
pounds  per  square  inch,  what  tension  will  rupture  a  bar  6  feet 
long  which  weighs  60  pounds? 

Prob.  6.  If  a  bar  I  inch  in  diameter  and  8  feet  long  elon- 
gates 0.05  inch  under  a  stress  of  15000  pounds,  how  much, 
according  to  law  (B),  will  a  bar  of  the  same  size  and  material 
elongate  whose  length  is  12  feet  and  stress  30000  pounds? 

ART.  4.     ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY. 

The  '  elastic  limit '  is  that  unit-stress  at  which  the  permanent 
set  is  first  visible  and  within  which  the  stress  is  directly  propor- 
tional to  the  deformation.  For  stresses  less  than  the  elastic 
limit  bodies  are  perfectly  elastic,  resuming  their  original  form 
on  removal  of  the  stress.  Beyond  the  elastic  limit  a  permanent 
alteration  of  shape  occurs,  or,  in  other  words,  the  elasticity  of 
the  material  has  been  impaired.  It  is  a  fundamental  rule  in  all 
engineering  constructions  that  materials  can  not  safely  be 
strained  beyond  their  elastic  limit. 

The  '  coefficient  of  elasticity  '  of  a  bar  for  tension,  compres- 
sion, or  shearing,  is  the  ratio  of  the  unit-stress  to  the  unit- 


8  RESISTANCE  AND   ELASTICITY   OF   MATERIALS.      CH.  I. 

deformation,  provided  the  elastic  limit  of  the  material  be  not 
exceeded.  Let  5  be  the  unit-stress,  s  the  unit-deformation,  and 
E  the  coefficient  of  elasticity.  Then  by  the  definition, 

(2)  E  =  -          and        5  =  Es. 

By  law  (B)  the  quantity  E  is  a  constant  for  each  material,  until 
S  reaches  the  elastic  limit.  Beyond  this  limit  s  increases  more 
rapidly  than  S  and  the  ratio  is  no  longer  constant.  Equation 
(2)  is  a  fundamental  one  in  the  science  of  the  strength  of 
materials.  Since  E  varies  inversely  with  s,  the  coefficient  of 
elasticity  may  be  regarded  as  a  measure  of  the  stiffness  of  the 
material.  The  stiffer  the  material  the  less  is  the  change  in 
length  under  a  given  stress  and  the  greater  is  E.  The  values 
of  E  for  materials  have  been  determined  by  experiments  with 
testing  machines  and  their  average  values  will  be  given  in  the 
following  articles.  E  is  necessarily  expressed  in  the  same  unit 
as  the  unit-stress  S.  Some  authors  give  the  name  '  modulus  of 
elasticity*  to  the  quantity  E. 

Another  definition  of  the  coefficient  of  elasticity  for  the  case 
of  tension  is  that  it  is  the  unit-stress  which  would  elongate  a 
bar  to  double  its  original  length,  provided  that  this  could  be 
done  without  exceeding  the  elastic  limit.  That  this  defini- 
tion is  in  agreement  with  (2)  may  be  shown  by  regarding  a  bar 
of  length  /  which  elongates  the  amount  A  under  the  unit- 

P  \ 

stress  -j.     Here  the  unit-elongation  is  7  and  (2)  becomes, 

P       A       PI 
(20  £-_=-+-=-, 

p 

and  if  \  be  equal  to  /,  E  is  the  same  as  the  unit-stress  -r. 

A 

Prob.  7.  Find  the  coefficient  of  elasticity  of  a  bar  of  wrought 
iron  \\  inches  in  diameter  and  16  feet  long  which  elongates  \ 
inch  under  a  tensile  stress  of  21  ooo  pounds. 


ART.  5. 


TENSION. 


Prob.  8.  If  the  coefficient  of  elasticity  of  cast  iron  is  1 5  ooo  ooo 
pounds  per  square  inch,  how  much  will  a  bar  2X3  inches  and 
6  feet  long  stretch  under  a  tension  of  5  ooo  pounds  ? 

ART.  5.    TENSION. 

The  phenomena  of  tension  observed  when  a  gradually  in- 
creasing stress  is  applied  to  a  bar,  are  briefly  as  follows  :  When 
the  unit-stress  .V  is  less  than  the  elastic  limit  5,,  the  unit-elon- 
gation s  is  small  and  proportional  to  5.  Within  this  limit  the 
ratio  of  5  to  s  is  the  coefficient  of  elasticity  of  the  material. 
After  passing  the  elastic  limit  the  bar  rapidly  elongates  and 
this  is  accompanied  by  a  reduction  in  area  of  its  cross-section. 
Finally  when  5  reaches  the  ultimate  tensile  strength  S(,  the 
bar  tears  apart.  Usually  St  is  the  maximum  unit-stress  on  the 
bar,  but  in  some  cases  the  unit-stress  reaches  a  maximum 
shortly  before  rupture  occurs. 

The  constants  of  tension  for  timber,  cast  iron,  wrought  iron 
and  steel  are  given  in  the  following  table.  The  values  are 
average  ones  and  are  liable  to  great  variations  for  different 
grades  and  qualities  of  materials.  Brick  and  stone  are  not 
here  mentioned,  as  they  are  rarely  or  never  used  in  tension. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Tensile 
Strength,  St  . 

Ultimate 
Elongation. 

Pounds  per  square 
inch. 

Pounds   per 
square  inch. 

Pounds  per 
square    inch. 

Per  cent. 

Timber, 

I  500000 

3  000 

IO  OOO 

1-5 

Cast  Iron, 

15  ooo  ooo 

6  000 

20  000 

o-5 

Wrought  Iron, 

25  oooooo 

25  ooo 

55  ooo 

20 

Steel, 

30  ooo  ooo 

50  ooo 

100  000 

IO 

The  values  of  the  coefficients  of  elasticity,  elastic  limits,  and 
breaking  or  ultimate  strengths  are  given  in  pounds  per  square 
inch  of  the  original  cross-section  of  the  bar.  The  ultimate  elon- 
gations are  given  in  percentages  of  the  original  length ;  these 


10 


RESISTANCE   AND   ELASTICITY   OF   MATERIALS.       CH.  I. 


express  elongations  per  linear  unit;  they  should  be  regarded 
as  very  rough  averages,  which  are  subject  to  great  variations 
depending  on  the  shape,  size,  and  quality  of  the  specimen. 

The  ultimate  elongation,  together  with  the  reduction  in  area 
of  the  cross-section,  furnishes  the  means  of  judging  of  the  duc- 
tility of  the  material.  The  reduction  of  area  in  cast  iron  and  in 
hard  steel  is  very  small,  while  in  ductile  materials  like  wrought 
iron  and  soft  steel  it  may  be  as  high  as  50  or  60  per  cent  of 
the  original  section. 

A  graphical  illustration  of  the  principal  phenomena  of  tension, 
is  given  in  Fig.  I.  The  unit-stresses  are  taken  as  ordinates  and 

100,0001 


0390,000 
•g  80,000 

a 

§  70,000 

"60,000 

I 

JS  50,000 

p,  40,000 

a 

S  30,000 

|  20,000 

10,000 

0 


Elongations  per  unit  of  length=  s 


:ht  Iron 


0.02 


0.04 


0.06  0.08 

Fig.  i. 


0.10 


0.12 


0.14 


0.16 


the  unit-elongations  as  abscissas.  For  each  unit-stress  the  cor- 
responding unit-elongation  as  found  by  experiment  is  laid  off, 
and  curves  drawn  through  the  points  thus  determined.  The 
curve  for  each  of  the  materials  is  a  straight  line  from  the  origin 
until  the  elastic  limit  is  leached,  as  should  be  the  case  accord- 
ing to  the  law  (S).  The  tangent  of  the  angle  which  this  line 
makes  with  the  axis  of  abscissas  is  equal  to  S  -=-  s,  which  is  the 
same  in  value  as  the  coefficient  of  elasticity  of  the  material. 
At  the  elastic  limit  a  sudden  change  in  the  curve  is  noticed  and 
the  elongation  rapidly  increases.  The  termination  of  the  curve 


ART.  5. 


TENSION. 


II 


indicates  the  point  of  rupture.  These  curves  show  more 
plainly  to  the  eye  than  the  values  in  the  table  can  do  the 
differences  in  the  properties  of  the  materials.  It  will  be  seen 
that  the  elastic  limit  is  not  a  well  defined  point,  but  that  its 
value  is  more  or  less  uncertain,  particularly  for  cast  iron  and 
timber.  It  should  be  also  clearly  understood  that  individual 
curves  for  special  cases  would  often  show  marked  variations 
from  their  mean  forms  as  represented  in  the  diagram. 

As  a  particular  example  a 
tensile  test  of  a  wrought  iron 
bar  f  inches  in  diameter  and 
12  inches  long  made  at  the 
Pencoyd  Iron  Works  will  be 
considered.  In  the  first  col- 
umn of  the  following  table 
are  given  the  total  stresses 
which  were  successively  ap- 
plied, in  the  second  the 
stresses  per  square  inch,  in 
the  third  the  total  elonga- 
tions, and  in  the  fourth  the 
elongations  or  sets  after  re- 
moval of  the  stress.  The 
unit-elongations  are  found  by 
dividing  those  in  the  table 
by  12  inches,  the  length  of 
the  specimen.  Then  the 
coefficient  of  elasticity  can 
be  computed  for  different 
values  of  .$  and  s.  Thus  for  the  fourth  and  seventh  cases, 


Total 
Stress 
in 
Pounds. 

Stress 
per 
Square 
Inch. 

Elongation. 

Load 
on. 

Load 
off. 

2  245 

5  000 

.001 

.000 

4490 

10  000 

.004 

.000 

6  735 

15  ooo 

.005 

.000 

8  980 

2O  OOO 

.008 

.000 

9878 

22  OOO 

.009 

.000 

10  776 

24  ooo 

.OIO 

.000 

ii  674 

26  ooo 

.0105 

.000 

12  572 

28  ooo 

.Oil 

.000 

13470 

30000 

.013 

.000 

14368 

32  000 

.014 

.000 

15  266 

34  ooo 

.015 

.002 

16  164 

36  ooo 

.022 

.007 

17  062 

38  ooo 

.416 

•3995 

17  960 

40  ooo 

•5445 

•523 

25  450 

50  ooo 

1.740 

1.707 

21  17^ 

CT    f»OO 

2  j68 

~J    */D                  *;  -    ~~~ 

Specimen  broke  with 

51  600  pounds 

per  square  inch. 

Stretch  in  12  inches,  2  468  inches. 

Stretch  in  8  inches,  1.812  inches. 

Stretch  in  8  inches,  22.65  Per  cent. 

Fractured  area,  o  297  square  inches. 

for  S  =  20  ooo,    s  — 


0.008 

12 

0.0105 

for  S  =  26  ooo,    s  =  - 

12 


and     £  =  30000000; 
and    E  =  29  700  ooo. 


12  RESISTANCE   AND    ELASTICITY    OF   MATERIALS.       LH.  1. 

The  elastic  limit  was  reached  at  about  33  ooo  pounds  per 
square  inch,  indicated  by  the  beginning  of  the  set  and  the  rapid 
increase  of  the  elongations.  The  ultimate  tensile  strength  of 
the  specimen  was  5 1  600  pounds  per  square  inch.  The  ultimate 
unit-elongation  in  8  inches  of  the  length  was  0.226  inches  per 
linear  inch.  It  hence  appears  that  this  bar  of  wrought  iron  was 
higher  than  the  average  as  regards  stiffness,  elastic  limit  and 
ductility,  and  lower  than  the  average  in  ultimate  strength. 

The  '  working  stress  '  for  a  material  is  that  unit-stress  to 
which  it  is,  or  is  to  be,  subjected.  This  should  not  be  greater 
than  the  elastic  limit  of  the  material,  since  if  that  limit  be  ex- 
ceeded there  is  a  permanent  set  which  impairs  the  elasticity. 
In  order  to  secure  an  ample  margin  of  safety  it  is  customary  to 
take  the  working  stresses  at  from  one-third  to  two-thirds  the 
"elastic  limit  Se.  The  reasons  which  govern  the  selection  of 
proper  values  of  the  working  stresses  will  be  set  forth  in  the 
following  articles. 

To  investigate  the  security  of  a  piece  subjected  to  a  tension 
Pt  it  is  necessary  first  to  divide  P  by  the  area  of  the  cross-sec- 
tion and  thus  determine  the  working  stress.  Then  a  com- 
parison of  this  value  with  the  value  of  Se  for  the  given  material 
will  indicate  whether  the  applied  stress  is  too  great  or  whether 
the  piece  has  a  margin  of  safety.  For  example,  if  a  tensile 
stress  of  4  500  pounds  be  applied  to  a  wrought  iron  bar  of  f 
inches  diameter  the  working  unit-stress  is, 

P      4500 

o  —  -j=  —      -  =  10  ooo  pounds  per  square  inch,  nearly. 
A       0.442 

As  this  is  less  than  one-half  the  elastic  limit  of  wrought  iron  the 
bar  has  a  good  margin  of  security. 

To  design  a  piece  to  carry  a  given  tension  P  it  is  necessary 
to  assume  the  kind  of  material  to  be  used  and  its  allowable 

p 
^working   stress   S.     Then   -^   is   the  area  of  the  cross-section 


ART.  6.  COMPRESSION.  13 

of  the  piece,  which  may  be  made  of  such  shape  as  the  circum- 
stances of  the  case  require.  For  example,  if  it  be  required  to 
design  a  wooden  bar  to  carry  a  tensile  load  of  4  500  pounds, 
the  working  stress  may  be  assumed  at  I  ooo  pounds  per 
square  inch  and  the  required  area  is  4.5  square  inches,  so  that 
the  bar  may  be  made  2  X  2j  inches  in  section. 

The  elongation  of  a  bar  within  the  elastic  limit  may  be  com- 
puted by  the  help  of  formula  (2).  For  instance,  let  it  be  re- 
quired to  find  the  elongation  of  a  wooden  bar  3X3  inches  and 
12  feet  long  under  a  tensile  stress  of  9000  pounds.  From  the 
formulas  (2)  and  (i), 

SP       X  Pl 


Substituting  in  this  the  values  E  =  i  500000,  A  =  g,  t=  144, 
and  P  =  9  ooo,  the  probable  value  of  the  elongation  X  is  found 
to  be  0.096  inches. 

Prob.  9.  Find  the  size  of  a  square  wooden  bar  to  safely 
carry  a  tensile  stress  of  20000  pounds. 

Prob.  10.  Compute  the  elongation  of  a  wrought-iron  bar, 
ii  inches  in  diameter  and  24  feet  long,  under  a  tension  of 
10000  pounds.  Also  that  of  a  cast-iron  bar. 

ART.  6.    COMPRESSION. 

The  phenomena  of  compression  are  similar  to  those  of  ten- 
sion, provided  that  the  length  of  the  specimen  does  not  exceed 
about  five  times  its  least  diameter.  The  piece  at  first  shortens 
proportionally  to  the  applied  stress,  but  after  the  elastic  limit  is 
passed  the  shortening  increases  more  rapidly,  and  is  accom- 
panied by  a  slight  enlargement  of  the  cross-section.  When  the 
stress  reaches  the  ultimate  strength  of  the  material  the  specimen 
cracks  and  ruptures.  If  the  length  of  the  piece  exceeds  about 
ten  times  its  least  diameter,  a  sidewise  bending  or  flexure  of 
the  specimen  occurs,  so  that  it  fails  under  different  circum- 


14  RESISTANCE   AND    ELASTICITY    OF   MATERIALS.       CH.  I. 


stances  than  those  of  direct  compression.  All  the  values  given 
in  this  article  refer  to  specimens  whose  lengths  do  not  exceed 
about  five  times  their  least  diameter.  Longer  pieces  will  be 
discussed  in  Chapter  V  under  the  head  of  'columns.'  Owing 
to  the  difficulty  of  making  experiments  on  short  specimens,  the 
phenomena  of  compression  are  not  usually  so  regular  as  those 
of  tension. 

The  constants  of  compression  for  short  specimens  are  given 
in  the  following  table,  the  values,  like  those  for  tension,  being 
rough  average  values  liable  to  much  variation  in  particular 
cases. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Compressive 
Strength,  Sc. 

Timber, 

Lbs.  per  sq.  in. 

i  500  ooo 

Lbs.  per  sq.  in. 
3  ooo 

Lbs.  per  sq.  in. 
8  000 

Brick, 

2  500 

Stone, 

6  OOO  OOO 

6  ooo 

Cast  Iron, 

15  ooo  ooo 

20  OOO 

90  ooo 

Wrought  Iron, 

25  ooo  ooo 

25  ooo 

55  ooo 

Steel, 

30  ooo  ooo 

50  ooo 

150  ooo 

The  values  of  the  coefficient  of  elasticity  and  the  elastic  limit 
for  timber,  wrought  iron,  and  steel  here  stated  are  the  same  as 
those  for  tension,  but  the  same  reliance  cannot  be  placed  upon 
them,  owing  to  the  irregularity  of  experiments  thus  far  made. 
There  is  reason  to  believe  that  both  the  elastic  limit  and  the 
coefficient  of  elasticity  for  compression  are  somewhat  greater 
than  for  tension. 

The  investigation  of  a  piece  subjected  to  compression,  or  the 
design  of  a  short  piece  to  be  subjected  to  compression,  is 
effected  by  exactly  the  same  methods  as  for  tension.  Indeed 
it  is  customary  to  employ  these  methods  for  cases  where  the 
length  of  the  piece  is  as  great  as  ten  times  its  least  diameter. 


ART.  7. 


SHEAR. 


Prob.  ii.  Find  the  height  of  a  brick  tower  which  crushes 
under  its  own  weight.  Also  the  height  of  a  stone  tower. 

Prob.  12.  Compute  the  amount  of  shortening  in  a  wrought 
iron  specimen  I  inch  in  diameter  and  5  inches  long  under  a  load 
of  6000  pounds. 

ART.  7.    SHEAR. 

Shearing  stresses  are  developed  whenever  two  forces,  act- 
ing like  a  pair  of  shears,  tend  to  cut  a  body  between  them. 
When  a  plate  is  punched  the  ultimate  shearing  strength  of  the 
material  must  be  overcome  over  the  surface  punched.  When 
a  bolt  is  in  tension  the  applied  stress  tends  to  shear  off  the 
head  and  also  to  strip  or  shear  the  threads  in  the  nut  and  screw. 
When  a  rivet  connects  two  plates  which  transmit  tension  the 
plates  tend  to  shear  the  rivet  across. 

The  ultimate  shearing  strength  of  materials  is  easily  deter- 
mined by  causing  rupture  under  a  stress  P,  and  then  dividing 
P  by  the  area  A  of  the  shorn  surface.  The  value  of  this  for 
timber  is  found  to  be  very  much  smaller  along  the  grain  than 
across  the  grain  ;  for  the  first  direction  it  is  sometimes  called 
longitudinal  shearing  strength  and  for  the  second  transverse 
shearing  strength.  The  same  distinction  is  sometimes  made 
in  rolled  wrought  iron  plates  and  bars  where  the  process  of 
manufacture  induces  a  more  or  less  fibrous  structure.  The 
•elastic  limit  and  the  amount  of  detrusion  for  shearing  are  dif- 


Material. 

Coefficient  of 
Elasticity,  £f 

Ultimate 
Shearing 
Strength,  St  . 

Timber,  Longitudinal, 

40OOOO 

600 

Timber,  Transverse, 

3000 

Cast  Iron, 

6000000 

20  OOO 

Wrought  Iron, 

10  000  OOO 

50000 

Steel, 

II  000000 

700OO 

16  RESISTANCE   AND    ELASTICITY   OF   MATERIALS.        CH.  L 

ficult  to  determine  experimentally.  The  coefficient  of  elas- 
ticity, however,  has  been  deduced  by  means  of  certain  calcula- 
tions and  experiments  on  the  twisting  of  shafts,  explained  in 
Chapter  VI  under  the  head  of  torsion. 

The  investigation  and  design  of  a  piece  to  withstand  shear- 
ing stress  is  made  by  means  of  the  equation  P  =  AS,  in 
the  same  manner  as  for  tension  and  compression.  As  an 

instance  of  investigation, 
consider  the  cylindrical 
wooden  specimen  shown 
in  Fig.  2,  which  has  the 

following  dimensions :  length  ab  =  6  inches,  diameter  of  ends 
=  4  inches,  diameter  of  central  part  =  2  inches.  Let  this 
specimen  be  subjected  to  a  tensile  stress  in  the  direction  of  its 
length.  This  not  only  tends  to  tear  it  apart  by  tension,  but 
also  to  shear  off  the  ends  on  a  surface  whose  length  is  ab  and 
whose  diameter  is  that  of  the  central  cylinder.  The  force  P 
required  to  cause  this  longitudinal  shearing  is, 

P  =  AS,  =  3.14  X  2  X  6  X  600  =  22  600  pounds, 

while  the  force  required  to  rupture  the  specimen  by  tension  is, 

P—  ASt  =  3.14  X  i2  X  10 ooo  =  31  400  pounds. 

As  the  former  resistance  is  only  about  two-thirds  that  of  the 
latter  the  specimen  will  evidently  fail  by  the  shearing  off  of 
the  ends. 

When  a  bar  is  subject  either  to  tension  or  to  compression 
a  shear  occurs  in  any  section  except  those  perpendicular  and 

parallel  to  the  axis  of  the  bar. 
Let  Fig.  3  represent  a  bar  of 
cross-section  A  subject  to  the 

tensile  stress  P  which  produces 
7 

Fig.  3.  in    every  section    perpendicular 

to  the  bar  the  unit-stress  —  .      Let  mn  be  a  plane  making  an* 

A 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.       17 

angle  6  with  the  axis,  and  cutting  from  the  bar  a  section  whose 
area  is  Al .  On  the  left  of  the  plane  the  stress  P  may  be  resolved 
'nto  the  components  Pl  and  Pt,  respectively  parallel  and  nor- 
mal to  the  plane,  and  the  same  may  be  done  on  the  right. 
Thus  it  is  seen  that  the  effect  of  the  tensile  stress  P  on  the 
plane  mn  is  to  produce  a  tension  P^  normal  to  it,  and  a  shear 
P,  along  it,  for  the  two  forces  Pl  and  Pl  act  in  parallel  planes 
and  in  opposite  directions.  The  shearing  stress  Pt  has  the 
value  P  cos  6,  which  is  distributed  over  the  area  Al  whose 
value  is  A  -+-  sin  0.  Hence  the  shearing  unit-stress  in  the 
given  section  is, 

P        P 

Sl=-±  =  -fs 
/i ,       A 

When  6  =  o°,  or  6  =  90°,  the  value  of  5X  is  zero.  The  maxi- 

p 

mum  value  of  5,  occurs  when  0  =  45°,  and  then  5,  =  J  — ,  or 

A 

a  tensile  unit-stress  5  on  a  bar  produces  a  shearing  unit-stress 
of  $S  along  every  section  inclined  45  degrees  to  the  axis  of  the 
bar.  The  above  investigation  applies  also  to  compression  if 
the  direction  of  P  be  reversed,  and  it  is  sometimes  observed  in 
experiments  on  the  compression  of  short  specimens  that  rup- 
ture occurs  by  shearing  along  oblique  sections. 

Prob.  13.  A  hole  }  inches  in  diameter  is  punched  in  a  wrought 
iron  plate  J  inches  thick  by  a  pressure  on  the  punch  of  78  ooo 
pounds.  What  is  the  ultimate  shearing  strength  of  the  iron? 

Prob  14.  A  wrought  iron  bolt  i£  inches  in  diameter  has  a 
head  £  inches  long.  Find  the  unit-stress  tending  to  shear  off 
ihe  head  when  a  tension  of  3  ooo  pounds  is  applied  to  the  bolt. 

ART.  8.    FACTORS  OF  SAFETY  AND  WORKING  STRESSES. 

The  factor  of  safety  for  a  body  under  stress  is  the  ratio  of  its 
ultimate  strength  to  the  actual  existing  unit-stress.  The  factor 
of  safety  for  a  piece  to  be  designed  is  the  ratio  of  the  ultimate 


i8 


RESISTANCE   AND    ELASTICITY   OF   MATERIALS.      CH.  I. 


strength  to  the  proper  allowable  working  stress.  Thus  if  St 
be  the  ultimate,  5  the  working  stress,  and  /  the  factor  of 
safety,  then 

/  =  •§,      and      St=fS. 

The  factor  of  safety  is  hence  always  an  abstract  number,  which 
indicates  the  number  of  times  the  working  stress  may  be  mul- 
tiplied before  the  rupture  of  the  body. 

The  law  (E)  in  Art.  3  indicates  that  working  stresses  should 
be  lower  for  shocks  and  sudden  stresses  than  for  steady  loads 
and  slowly  varying  stresses.  In  a  building  the  stresses  on  the 
walls  are  steady,  so  that  the  working  strength  may  be  taken 
high  and  hence  the  factor  of  safety  low.  In  a  bridge  the 
stresses  in  the  several  members  are  more  or  less  varying  in 
character  which  requires  a  lower  working  strength  and  hence 
a  higher  factor  of  safety.  In  a  machine  subject  to  shocks  the 
working  strength  should  be  lower  still  and  the  factor  of  safety 
very  high.  The  law  (E)  from  which  these  conclusions  are  de- 
rived is  not  merely  the  result  of  experience,  but  can  be  con- 
firmed by  theoretical  discussion  (Art.  103). 

The  following  are  average  values  of  the  allowable  factors  of 
safety  commonly  employed  in  American  practice.  These  values 


Material. 

For  Steady 
Stress. 
(Buildings.) 

For  Varying 
Stress. 
(Bridges.) 

For  Shocks. 
(Machines.) 

Timber, 

8 

IO 

15 

Brick  and  Stone, 

15 

25 

30 

Cast  Iron, 

6 

15 

2O 

Wrought  Iron, 

4 

6 

10 

Steel, 

5 

7 

15 

are  subject  to  considerable  variation  in  particular  instances,  not 
only  on  account  of  the  different  qualities  and  grades  of  the 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.       19 

material,  but  also  on  account  of  the  varying  judgment  of 
designers.  They  will  also  vary  with  the  range  of  varying  stress, 
so  that  different  parts  of  a  bridge  may  have  very  different 
factors  of  safety. 

The  proper  allowable  working  stress  for  any  material  in 
tension,  compression,  or  shearing,  may  be  at  once  found  by 
dividing  the  ultimate  strength  by  the  proper  factor  of  safety. 
Regard  should  also  be  paid  to  the  elastic  limit  in  selecting  the 
working  stresses,  particularly  for  materials  whose  elastic  limit 
is  well  defined.  For  wrought  iron  and  steel  the  working 
strength  should  be  well  within  the  elastic  limit,  as  already  in- 
dicated in  previous  articles.  For  cast  iron,  stone,  brick,  and 
timber  it  is  often  difficult  to  determine  the  elastic  limit,  and 
experience  alone  can  guide  the  proper  selection  of  the  working 
strength.  The  above  factors  of  safety  indicate  indeed  the  con- 
clusions of  experiment  and  experience  extending  over  the  past 
hundred  years. 

The  student  should  clearly  understand  that  the  exact  values 
given  in  this  and  the  preceding  articles  would  not  be  arbitrarily 
used  in  any  particular  case  of  design.  For  instance,  if  a  given 
lot  of  wrought  iron  is  to  be  used  in  an  engineering  structure, 
specimens  of  it  should  be  tested  to  determine  its  coefficient  of 
elasticity,  elastic  limit,  ultimate  strength,  and  percentage  of 
elongation.  Then  the  engineer  will  decide  upon  the  proper 
working  stresses,  being  governed  by  its  qualities  as  shown  by 
the  tests,  the  character  of  the  stresses  that  come  upon  it,  and 
the  cost  of  workmanship. 

The  two  fundamental  principles  of  engineering  design  are 
stability  and  economy,  or  in  other  words : 

First,  the  structure  must  safely  withstand  all  the  stresses 

which  are  to  be  applied  to  it. 
Second,  the  structure  must  be  built  and  maintained  at  the 

lowest  possible  cost. 


20  RESISTANCE  AND   ELASTICITY   OF   MATERIALS.       CH.  1. 

The  second  of  these  fundamental  principles  requires  that  all 
parts  of  the  structure  should  be  of  equal  strength,  like  the 
celebrated  'one-hoss  shay'  of  the  poet.  For,  if  one  part  is 
stronger  than  another,  it  has  an  excess  of  material  which  might 
have  been  spared.  Of  course  this  rule  is  to  be  violated  if  the 
cost  of  the  labor  required  to  save  the  material  be  greater  than 
that  of  the  material  itself.  Thus  it  often  happens  that  some 
parts  of  a  structure  have  higher  factors  of  safety  than  others, 
but  the  lowest  factors  should  not,  as  a  rule,  be  less  than  the 
values  given  above.  .For  the  design  of  important  structures 
specifications  are  prepared  which  state  the  lowest  allowable 
unit-stresses  that  can  be  used. 

The  factors  of  safety  stated  above  are  supposed  to  be  so 
arranged  that,  if  different  materials  be  united,  the  stability  of 
all  parts  of  the  structure  will  be  the  same,  so  that  if  rupture 
occurs,  everything  would  break  at  once.  Or,  in  other  words, 
timber  with  a  factor  of  safety  8  has  about  the  same  reliability 
as  wrought  iron  with  a  factor  of  4  or  stone  with  a  factor  of  15, 
provided  the  stresses  are  due  to  steady  loads. 

The  assignment  of  working  stresses  with  regard  to  the 
elastic  limits  of  materials  is  more  rational  than  that  by  means 
of  the  factors  of  safety,  and  in  time  it  may  become  the  more 
important  and  valuable  method.  But  at  present  the  ultimate 
strengths  are  so  much  better  known  and  so  much  more  definitely 
determinable  than  the  elastic  limits  that  the  empirical  method 
of  factors  of  safety  seems  the  more  important  for  the  use  of 
students,  due  regard  being  paid  to  considerations  of  stiffness, 
elastic  limit,  and  ductility. 

As  an  example,  let  it  be  required  to  find  the  proper  size  of  a 
wrought  iron  rod  to  carry  a  steady  tensile  stress  of  90  ooo 
pounds.  In  the  absence  of  knowledge  regarding  the  quality 
of  the  wrought  iron,  the  ultimate  strength  St  is  to  be  taken  as 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.      21 

the  average  value,  55  ooo  pounds  per  square  inch.     Then,  for  a 
factor  of  safety  of  4,  the  working  stress  is, 

S  =  -  =13  750  pounds  per  square  inch. 

The  area  of  cross-section  required  is  hence, 

90000 
A =  6.6  square  inches, 

which  may  be  supplied  by  a  rod  of  2|$  inches  diameter. 

Prob.  15.  Determine  the  size  of  a  short  steel  piston  rod  when 
the  piston  is  20  inches  in  diameter  and  the  steam  pressure  upon 
it  is  67.5  pounds  per  square  inch. 

Prob.  16.  A  wooden  frame  ABC  forming  an  equilateral  tri- 
angle consists  of  short  pieces  2X2  inches  jointed  at  A,  B,  and 
C.  It  is  placed  in  a  vertical  plane  and  supported  at  B  and  C 
so  that  BC  is  horizontal.  Find  the  unit-stress  and  factor  of 
safety  in  each  of  the  three  pieces  when  a  load  of  5  890  pounds 
is  applied  at  A. 


22  PIPES,   CYLINDERS,   AND    RIVETED  JOINTS.        CH.  LL 


CHAPTER   II. 
PIPES,  CYLINDERS,   AND   RIVETED  JOINTS. 

ART.  9.    WATER  AND  STEAM  PIPES. 

The  pressure  of  water  or  steam  in  a  pipe  is  exerted  in  every 
direction,  and  tends  to  tear  the  pipe  apart  longitudinally.  This 
is  resisted  by  the  internal  tensile  stresses  of  the  material.  If 
/  be  the  pressure  per  square  inch  of  the  water  or  steam,  d  the 
diameter  of  the  pipe  and  /  its  length,  the  force  P  which  tends 
to  cause  longitudinal  rupture  is/ .  Id.  This  is  evident  from  the 
fundamental  principle  of  hydrostatics  that  the  pressure  of  water 
in  any  direction  is  equal  to  the  pressure  on  a  plane  perpen- 
dicular to  that  direction,  or  may  be  seen  by  imagining  the  pipe 
to  be  filled  with  a  solid  substance  on  one  side  of  the  diameter, 
which  would  receive  the  pressure  /  on  each  square  inch  of  the 
area  Id  and  transmit  it  into  the  pipe.  If  /  be  the  thickness  of 
the  pipe  and  5  the  tensile  stress  which  is  uniformly  distributed 
over  it,  as  will  be  the  case  when  /  is  not  large  compared  with 
d,  the  resistance  on  each  side  is  //.  5.  As  the  resistance  must 
equal  the  pressure, 

pld  —  2tlS,       or      pd  =  2/5, 
which  is  the  formula  for  discussing  pipes  under  internal  pressure. 

The  unit-pressure/  for  water  may  be  computed  from  a  given 
head  h  by  finding  the  weight  of  a  column  of  water  one  inch 
square  and  h  inches  high.  Or  if  h  be  given  in  feet,  the  pressure 
in  pounds  per  square  inch  may  be  computed  from  /  =  0.434^. 

Water  pipes  maybe  made  of  cast  or  wrought  iron,  the  former 
being  more  common,  while  for  steam  the  latter  is  preferable. 


ART.  9.  WATER  AND   STEAM    PIPES.  23 

Wrought  iron  pipes  are  sometimes  made  of  plates  riveted  to- 
gether, but  the  discussion  of  these  is  reserved  for  another 
article.  A  water  pipe  subjected  to  the  shock  of  water  ram 
needs  a  high  factor  of  safety,  and  in  a  steam  pipe  the  factors 
should  also  be  high,  owing  to  shocks  liable  to  occur  from  con- 
densation and  expansion  of  the  steam.  The  formula  above 
deduced  shows  that  the  thickness  of  a  pipe  must  increase 
directly  as  its  diameter,  the  internal  pressure  being  constant. 

For  example,  let  it  be  required  to  find  the  factor  of  safety 
for  a  cast  iron  water  pipe  of  12  inches  diameter  and  f  inches 
thickness  under  a  head  of  300  feet.  Here  /,  the  pressure  per 
square  inch,  equals  130.2  pounds.  Then  from  the  formula  the 
unit-stress  is, 

pd      130.2  X  12 
5  =  *—  =  -  —5  —  =  i  250  pounds  per  square  inch, 

21  2  X  •§• 

and  hence  the  factor  of  safety  is, 

20000 


which  indicates  ample  security  under  ordinary  conditions. 

Again  let  it  be  required  to  find  the  proper  thickness  for  a 
wrought  iron  steam  pipe  of  18  inches  diameter  to  resist  a  pres- 
sure of  1  20  pounds  per  square  inch.  With  a  factor  of  safety 
of  10  the  working  strength  5  is  about  5  500  pounds  per  square 
inch.  Then  from  the  formula, 

pd      120  X  18 

/  =  £—  =  —         -  =  0.2  inches. 
25      2  X  5  500 

In  order  to  safely  resist  the  stresses  and  shocks  liable  to  occur 
in  handling  the  pipes,  the  thickness  is  often  made  somewhat 
greater  than  the  formula  requires. 

Prob.  17.  What  should  be  the  thickness  of  a  cast  iron  pipe 
of  1  8  inches  diameter  under  a  head  of  300  feet? 

Prob.  1  8.  A  wrought  iron  pipe  is  3  inches  in  internal  diame- 


24  PIPES,   CYLINDERS,   AND    RIVETED   JOINTS.        CH.  II. 

ter  and  weighs  24  pounds  per  linear  yard.     What  steam  pres- 
sure can  it  carry  with  a  factor  of  safety  of  8  ? 


ART.  10.    THIN  CYLINDERS  AND  SPHERES. 

A  cylinder  subject  to  the  interior  pressure  of  water  or  steam 
tends  to  fail  longitudinally  exactly  like  a  pipe.  The  head  of 
the  cylinder  however  undergoes  a  pressure  which  tends  to 
separate  it  from  the  walls.  If  d  be  the  diameter  of  the  cylin- 
der and  /  the' internal  pressure  per  square  unit,  the  total  pres- 
sure on  the  head  is  ^nd* .  p.  If  5  be  the  working  unit-stress 
and  t  the  thickness  of  the  cylinder,  the  resistance  to  the  pres- 
sure is  approximately  ndtS,  if  t  be  so  small  that  5  is  uniformly 
distributed.  Since  the  resistance  must  equal  the  pressure, 

\nd*  .p  =  7tdt.S,         or        pd  —  tfS. 

By  comparing  this  with  the  formula  of  the  last  article  it  is  seen 
that  the  resistance  of  a  pipe  to  transverse  rupture  is  double  the 
resistance  to  longitudinal  rupture. 

A  thin  sphere  subject  to  interior  pressure  tends  to  rupture 
around  a  great  circle,  and  it  is  easy  to  see  that  the  conditions 
are  exactly  the  same  as  for  the  transverse  rupture  of  a  cylin- 
der, or  that  pd  —  4tS.  For  thick  spheres  and  cylinders  the 
formulas  of  this  and  the  last  article  are  only  approximate. 

A  cylinder  under  exterior  pressure  is  theoretically  in  a  simi- 
lar condition  to  one  under  interior  pressure  as  long  as  it  re- 
mains a  true  circle  in  cross-section.  A  uniform  interior  pres- 
sure tends  to  preserve  and  maintain  the  circular  form  of  the 
cylindrical  annulus,  but  an  exterior  pressure  tends  at  once  to 
increase  the  slightest  variation  from  the  circle  and  render  it 
elliptical.  The  distortion  when  once  begun  rapidly  increases, 
and  failure  occurs  by  the  collapsing  of  the  tube  rather  than  by 
the  crushing  of  the  material.  The  flues  of  a  steam  boiler  are 
the  most  common  instance  of  cylinders  subjected  to  exterior 


ART.  10. 


THIN   CYLINDERS   AND   SPHERES. 


pressure.  In  the  absence  of  a  rational  method  of  investigating 
such  cases  recourse  has  been  had  to  experiment.  Tubes  of 
various  diameters,  lengths,  and  thicknesses  have  been  subjected 
to  exterior  pressure  until  they  collapse  and  the  results  have 
been  compared  and  discussed.  The  following  for  instance  are 
the  results  of  three  experiments  by  FAIRBAIRN  on  wrought 
iron  tubes. 


Length 
in 
Inches. 

Diameter, 
in 
Inches. 

Thickness 
in 
Inches. 

Pressure 
per 
Sq.  Inch. 

37 

9 

0.14 

378 

60 

14* 

0.125 

125 

61 

iSf 

0.25 

420        , 

From  these  and  other  similar  experiments  it  has  been  concluded 
that  the  collapsing  pressure  varies  directly  as  some  power  of  the 
thickness,  and  inversely  as  the  length  and  diameter  of  the  tube. 
For  wrought  iron  tubes  WOOD  gives  the  empirical  formula  for 
the  collapsing  pressure  per  square  inch, 

/2.l8 

/  ==  9600000-—-. 

The  values  of  /  computed  from  this  formula  for  the  above 
three  experiments  are  397,  120,  and  409,  which  agree  well  with 
the  observed  values. 

The  proper  thickness  of  a  wrought  iron  tube  to  resist  ex- 
terior pressure  may  be  readily  found  from  this  formula  after 
assuming  a  suitable  factor  of  safety.  For  example,  let  it  be 
required  to  find  /  when  p  =  120  pounds  per  square  inch,  /  =  72 
inches,  d  =  4  inches  and  the  factor  of  safety  =  10.  Then 

,,..,=  10X.20X72>M=  g 

9600000 

from  which  with  the  help  of  logarithms  the  value  of  /  is  found 
to  be  0.22  inches. 


26  PIPES,   CYLINDERS,   AND   RIVETED   JOINTS.        CH.  IL 

Prob.  19.  What  interior  pressure  per  square  inch  will  burst 
a  cast  iron  sphere  of  24  inches  diameter  and  |  inches  thickness. 

Prob.  20.  What  exterior  pressure  per  square  inch  will  col- 
lapse a  wrought  iron  tube  72  inches  long,  4  inches  diameter  and 
0.25  inches  thickness? 

ART.  ii.    THICK  CYLINDERS. 

When  the  walls  of  a  cylinder  are  thick  compared  with  its 
interior  diameter  it  cannot  be  supposed,  as  in  the  preceding 
articles,  that  the  stress  is  uniformly  distributed  over  the  thick- 
ness /.  Let  Fig.  4  represent  one-half  of  a  section  of  a  thick 

cylinder  subject  to  interior  pressure 
over  the  length  /,  tending  to  produce 
longitudinal  rupture.  Let  r  and  rl 
be  the  interior  and  exterior  radii, 
then  T-,  —  r  =  t  the  thickness.  Let 
5  and  S,  be  the  tensile  unit-stresses 
at  the  inner  and  outer  edges  of  the 
annulus.  Before  the  application  of 
the  pressure  the  volume  of  the  annulus  is  n(r*  —  r2)/,  after  the 
pressure  is  applied  the  radius  rl  is  increased  to  rl  -\-  yl  and  r  to 
r  +  y,  so  that  its  volume  is  n(rl  +  y$l  —  n(r  +  yJL  The  an- 
nulus is  however  really  changed  only  in  form,  so  that  the  two 
expressions  for  the  volume  are  equal,  and  equating  them  gives,, 


or,  since  y  and  7,   are  small  compared  with  r  and  r,  their 
squares  may  be  neglected,  and  hence 


or  -  =  ~ 


Now  if  the  material  is  not  stressed  beyond  the  elastic  limit  the 
unit-stresses  6"  and  ^  are  proportional  to  the  corresponding 
unit-elongations.  The  elongation  of  the  inner  circumference 
is  27ty  and  that  of  the  outer  circumference  is  27tylt  and  divid- 


ART.  II.  THICK   CYLINDERS.  2/ 

ing  these  by  2nr  and  2nrl  respectively  the  unit-elongations  are 
found;  then, 

—  =  2  -*.**  =  ?*  .  L. 

S,       r   '    r,       r  '  y^ 
Substituting  in  this  the  value  of  the  ratio  —  as  above  found, 


that  is,  the  unit-stresses  in  the  walls  of  the  cylinder  .vary  in- 
versely as  the  squares  of  their  distances  from  the  center. 

The  total  stress  acting  over  the  area  2t  .  /  is  now  to  be  found 
by  summing  up  the  unit-stresses.  Let  Sx  be  any  unit-stress  at 
a  distance  x  from  the  center,  and  5,  as  before,  be  that  at  the 
inner  circumference,  which  is  the  greatest  of  all  the  unit-stresses* 
Then  by  the  law  of  variation, 


The  stress  acting  over  the  area  /.  dx  ?*s  then 

5,  Idx  =  SSl?jr, 
and  the  total  stress  over  the  area  2t  .  I  is 


25rV  r~,  = 
Jr    ** 


-       =  2SlJ* 


This  is  the  value  of  the  internal  resisting  stress  in  the  walls  of 
the  pipe  ;  if  /  be  neglected  in  comparison  with  r  it  reduces  to 
2Slt  which  is  the  same  as  previously  found  for  thin  cylinders  ; 
if  t  =  r  it  becomes  Sit  or  only  one-half  the  resistance  of  a  thin 
cylinder. 

The  total  interior  pressure  which  tends  to  rupture  the  cylin- 
der longitudinally  is  2rl  .  /,  if  p  be  the  unit-pressure  (Art.  9). 


28  PIPES,    CYLINDERS,   AND   RIVETED  JOINTS.        CH.  II. 

Equating  this  to  the  total  internal  resisting  stress  gives 


from  which  one  of  the  quantities  S,  p,  r,  or  /  can  be  computed 
when  the  other  three  are  given. 

The  above  formula  was  deduced  by  BARLOW.  Although 
more  accurate  for  thick  cylinders  than  the  formula  of  Art.  10, 
it  is  not  in  practice  considered  so  reliable  as  the  formula  of 
LAME  which  is  deduced  in  Chapter  XIV.  If  '  r  =  6  inches, 
/  —  i  inch,  and/  =  400  pounds  per  square  inch,  BARLOW'S 
formula  gives  5=2800,  while  LAME'S  formula  (page  314) 
gives  5  =  2620  pounds  per  square  inch. 

Prob.  21.  Prove  when  the  thickness  of  a  pipe  equals  its  in- 
terior radius  that  the  exterior  circumference  elongates  one-half 
as  much  as  the  interior  circumference. 

Prob.  22.  If  a  gun  of  3  inches  bore  is  subject  to  an  interior 
pressure  of  I  Soc  pounds  per  square  inch,  what  should  be  its 
thickness  so  that  the  greatest  stress  on  the  material  may  not 
exceed  3  ooo  pounds  per  square  inch  ? 

ART.  12.    INVESTIGATION  OF  RIVETED  JOINTS. 

When  two  plates  which  are  under  tension  are  joined  togethei 
by  rivets,  these  must  transfer  that  tension  from  one  plate  to 
another.  A  shearing  stress  is  thus  brought  upon  each  rivet 
which  tends  to  cut  it  off.  A  compressive  stress  is  also  brought 
sidewise  upon  each  rivet  which  tends  to  crush  it  ;  this  particu- 
lar kind  of  compression  is  often  called  "bearing  stress."  The 
exact  manner  in  which  it  acts  upon  the  cylindrical  surface  of 
the  rivet  is  not  known,  but  it  is  usually  supposed  to  be  equiva- 
lent to  a  stress  uniformly  distributed  over  the  projection  of  the 

surface  on  a  plane  through  the  axis  of  the  rivet. 

ft 
Case  I.  Lap  Joint  with  single  riveting.  —  Let  P  be  the  tensile 

stress  which  is  transmitted   from  one  plate  to  the  other  by 


ART.   12.         INVESTIGATION    OF   RIVETED   JOINTS. 


29 


means    of  a  single    rivet,  t  the  thickness   of  the  plates,  d  the 
diameter  of  a  rivet,  and  a  the  pitch  of  the  rivets.     Let  St,  Sft 
and   Sc  be    the    unit-stresses   in 
tension,  shear,  and  compression  ^N 

produced  by  P  upon  the  plates   "  s^ 

and   rivets.     Then    for  the   ten- 
sion on  the  plate,  -* — 

P  =  t(a-  d)St , 
for  the  shear  on  the  rivet, 


and  for  compression  on  the  rivet, 


Fig.  5- 


From  these  equations  the  unit-stresses  may  be  computed,  when 
the  other  quantities  are  known,  and  by  comparing  them  with 
the  proper  working  unit-stresses  the  degree  of  security  of  the 
joint  is  estimated. 

Case  II.  Lap  Joint  with  double  riveting.  —  In  this  arrange- 
ment the  plates  have  a  wider  lap, 
and  there  are  two  rows  of  rivets. 
Let  a  be  the  pitch  of  the  rivets  in 
one  row,  then  the  tensile  stress  P  is 
distributed  over  two  rivets,  and  the 
three  formulas  are, 

P=t(a-d}Sty 


\ 


from  which  the  unit-stresses  may  be 

computed    and    the   strength    of    the 

joint    be    investigated.     The    loss    of  Fig.  6 

strength  is  here  generally  less  than  in  the  previous  case  since  a 

can  be  made  larger  with  respect  to  d. 


30  PIPES,   CYLINDERS,    AND    RIVETED   JOINTS.        CH.  II. 

Case  III.  Butt  Joint  with  single  riveting. — For  this  arrange- 
^ — ^  /• — x.  ment  the  shear  on  each  of 

the    rivets    comes    on    two 

X^X  X^x  cross-sections,  which  is  said 

Fig.  7.  to  be  a  case  of  double  shear, 

and  the  formulas  are, 

P=t(a-  d)Sty  =  2  .  i-7r</2Ss  =  tdSe. 

Accordingly,  a  lap  joint  with  double  riveting  has  the  same  ten- 
sile and  shearing  strength  as  a  butt  joint  with  single  riveting, 
if  the  values  of  a,  d,  and  t  be  equal  in  the  two  cases ;  the  bear- 
ing resistance,  however,  is  only  one  half  as  large. 

For  example,  let  it  be  required  to  investigate  a  single  riveted 
butt  joint  consisting  of  plates  0.75  inches  thick  with  covers 
0.375  inches  thick,  and  rivets  of  I  inch  diameter  and  4  inches 
pitch,  when  a  tension  of  8  ooo  pounds  is  transmitted  through 
one  rivet.  First,  the  working  tensile  unit-stress  on  the  plate  is 
found  to  have  the  value, 

0          8000 

—  ~~^ — ^l  =  3  56°  Pounds  per  square  inch. 

Next  the  shearing  unit-stress  on  the  rivets  is, 

8000 
os  =  — — — -Q—  =  5  100  pounds  per  square  inch. 

2   X.  0.785 

Lastly,  the  bearing  compressive  unit-stress  on  the  rivets  is, 

8000 
Se  = =  10  700  pounds  per  square  inch. 

It  thus  appears  that  the  joint  has  the  greatest  factor  of  safety 
against  tension  and  the  least  against  compression  of  the  rivets. 
It  should  be  said,  however,  that  for  wrought  iron  plates  and 
rivets  the  highest  allowable  working  stresses  for  tension,  shear, 
and  bearing  are  generally  considered  to  be  about  9  ooo,  7  500, 
and  1 2  ooo  pounds  per  square  inch  respectively;  hence  the 


ART.  12.         INVESTIGATION   OF   RIVETED   JOINTS.  31 

joint  has  proper  security  under  the  given  conditions  although 
the  degree  of  security  is  quite  different  for  the  different 
stresses. 

The  'efficiency'  of  a  joint  is  defined  to  be  the  ratio  of  its 
highest  allowable  stress  to  the  highest  allowable  stress  of  the 
unriveted  plate.  The  highest  allowable  stresses  in  tension, 
shear,  and  compression  are  the  three  expressions  for  P,  using, 
for  wrought  iron,  the  values  above  mentioned  ;  and  the  highest 
allowable  stress  of  the  unriveted  plate  isatSt.  Thus  result  the 
following  values  of  the  efficiency, 

a-d 
For  tension,  e  =  -      -  » 


For  shear, 


.                           M  .  dSc 
For  compression,  e  = ^ —  » 

in  which  m  denotes  the  number  of  rivets  in  the  width  a  which 
transmit  the  tension  P  and  n  denotes  the  number  of  rivet-sec- 
tions  in  the  same  space  over  which  the  shear  is  distributed. 
The  smallest  of  these  values  of  e  is  to  be  taken  as  the  efficiency 
of  the  joint.  Thus  for  the  above  numerical  example  the  three 
values  are  0.75,  0.44,  and  0.33 ;  the  working  strength  of  the 
joint  is,  hence,  only  33  per  cent  of  that  of  the  unriveted  plate. 

If  in  the  above  formulas,  St,  St,  and  Sc  be  taken  as  the  ulti- 
mate strengths,  the  resulting  values  of  e  will  be  the  efficiencies 
at  the  moment  of  rupture  of  the  joint.  For  the  same  numerical 
example  the  three  ultimate  efficiencies  are  0.75,  0.48,  and  0.25. 

Prob.  23.  A  boiler  is  to  be  formed  of  wrought  iron  plates  f 
inches  thick,  united  by  single  lap  joints  with  rivets  f  inches 
diameter  and  if  inches  pitch.  Find  the  efficiency  of  the  joint. 
Find  the  factor  of  safety  of  the  boiler  if  it  is  30  inches  in  di- 
ameter and  carries  a  steam  pressure  of  100  pounds. 


32  PIPES,  CYLINDERS,  AND  RIVETED  JOINTS.      CH.  IL 

ART.  13.    DESIGN  OF  RIVETED  JOINTS. 

A  theoretically  perfect  joint  with  regard  to  strength  is  one 
so  arranged  that  all  parts,  (like  the  one-hoss  shay)  have  the 
same  degree  of  security.  Thus  the  resistance  of  the  plate  to 
tension  must  equal  the  resistance  of  the  rivets  to  shearing,  and 
each  of  these  must  equal  the  resistance  of  the  rivets  to  com- 
pression. The  three  expressions  for  P  of  the  last  Article  should 
hence  be  equal,  or,  what  amounts  to  the  same  thing,  the  three 
efficiencies  should  be  equal.  Equating  then  the  second  to  the 
third  and  solving  for  </,  gives 


from  which  d  can  be  computed  when  t  is  assumed.     Again, 
equating  the  first  and  third  and  eliminating  d  gives, 

,  ™s; 


from  which  the  pitch  of  the  rivets  can  be  obtained.  Inserting 
these  values  of  d  and  a  in  either  of  the  expressions  for  e  fur- 
nishes the  formula, 


e  =  - 


rmSe 

from  which  the  efficiency  can  be  ascertained.  The  best  'joint 
will  be  that  which  has  the  least  loss  of  strength  due  to  the 
riveting,  or  that  which  has  a  value  of  e  as  near  unity  as  possible. 

Using  for  wrought  iron  plates  and  rivets  the  working  unit- 
stresses  St  =  9000,  5S  =  7  500,  and  Sc  —  12000  pounds  per 
square  inch,  the  above  formulas  for  a  lap  joint  with  single  row 
of  rivets  where  m  =  I  and  n  =  I,  reduce  to, 

d  =  2.04*,  a  =  4.75*,  e  =  0.57, 

so  that,  if  the  thickness  of  the  plate  be  given  and  the  diameter 
and  pitch  of  the  rivet  be  made  according  to  these  rules,  the 


ART.  13.  DESIGN   OF  RIVETED   JOINTS.  33 

joint  has  about  57  per  cent  of  the  strength  of  the  unholed 
plate.  For  a  lap  joint  with  double  riveting  where  m  =  2  and 
n  =  2,  they  become 

d  —  2.04/,  a  =  7.48*,  e  =  0.73. 

This  example  shows  clearly  the  advantage  of  double  over  single 
riveting,  and  by  adding  a  third  row  the  efficiency  will  be  raised 
to  about  80  per  cent. 

The  application  of  the  above  formulas  to  butt  joints  makes 
the  diameter  of  the  rivet  equal  to  the  thickness  of  the  plate 
and  makes  the  pitch  much  smaller  than  the  above  values  for 
lap  joints.  These  proportions  are  difficult  to  apply  in  practice 
on  account  of  the  danger  of  injuring  the  metal  in  punching  the 
holes.  For  this  reason  joints  are  often  made  in  which  the 
strengths  of  the  different  parts  are  not  equal.  Many  other 
reasons,  such  as  cost  of  material  and  facility  of  workmanship, 
influence  also  the  design  of  a  joint  so  that  the  formulas  above 
deduced  are  to  be  regarded  only  as  a  rough  guide.  The  old 
rules  which  are  still  often  used  for  determining  the  pitch  in  butt 
joints,  are 


the  first  being  for  single  and  the  second  for  doubje  riveting. 
These  are  deduced  by  making  the  strength  of  the  joint  equal 
in  tension  and  shear,  and  taking  Ss  =  St. 

It  may  be  required  to  arrange  a  joint  so  as  to  secure  eithei 
strength  or  tightness.  For  a  bridge,  strength  is  mainly  needed  ; 
for  a  gasholder,  tightness  is  the  principal  requisite;  while  for  a 
boiler  both  these  qualities  are  desirable.  In  general  a  tight 
joint  is  secured  by  using  small  rivets  with  a  small  pitch.  The 
lap  of  the  plates,  and  the  distance  between  the  rows  of  rivets, 
is  determined  by  practical  considerations  rather  than  by 
theoretic  formulas. 


34  PIPES,    CYLINDERS,    AND    RIVETED   JOINTS.        CH.  II. 

Prob.  24.  A  lap  joint  with  double  riveting  is  to  be  formed 
of  plates  %  inches  thick  with  rivets  -J  inches  diameter.  Find 
the  pitch  so  that  the  strength  of  the  plate  shall  equal  the 
shearing  strength  of  the  rivets,  and  compute  the  efficiency  of 
the  joint. 

ART.  14.    MISCELLANEOUS  EXERCISES. 

It  will  be  profitable  to  the  student  to  thoroughly  perform  the 
following  exercises  and  problems  and  to  write  upon  each  a 
detailed  report,  which  should  contain  all  the  sketches  and 
computations  necessary  to  clearly  explain  the  data,  the  reason- 
ing, the  computations,  and  the  conclusions.  Problem  26  is 
intended  for  students  proficient  in  the  use  of  calculus. 

Exercise  I.  Visit  an  establishment  where  tensile  tests  are 
made.  Ascertain  the  kind  of  machine  employed,  its  capacity, 
the  method  of  applying  the  stresses,  the  method  of  measuring 
the  stresses,  the  method  of  measuring  the  elongations.  Ascer- 
tain the  kind  of  material  tested,  the  reason  for  testing  it,  and 
the  conclusions  derived  from  the  tests.  Give  full  data  for  the 
tests  on  four  different  specimens,  compute  the  values  of  co- 
efficient of  elasticity,  ultimate  strength  and  ultimate  elongation 
for  them,  and  state  your  conclusions. 

Exercise  2.  Procure  a  wrought  iron  bolt  and  nut.  Measure 
the  diameter  of  bolt,  length  of  head,  and  length  of  nut.  State 
the  equation  of  condition  that  the  head  of  the  bolt  may  shear 
off  at  the  same  time  the  bolt  ruptures  under  tension.  Com- 
pute the  length  of  head  for  a  given  diameter.  Explain  why 
the  length  of  the  head  is  made  greater  than  theory  apparently 
requires.  Compile  a  table  giving  dimensions  of  bolts  and  nuts 
of  different  diameters. 

Exercise  3.  Go  to  a  boiler  shop  and  witness  operations  upon  a 
boiler  in  process  of  construction.  Ascertain  length  and  diameter 
of  boiler,  thickness,  pitch  and  diameter  of  rivets,  method  of 
forming  holes,  method  of  doing  the  riveting.  Compute  the  loss 
of  strength  caused  by  the  riveting.  Compute  the  steam  pressure 


ART.   14.  MISCELLANEOUS    EXERCISES.  35 

which  would  cause  longitudinal  rupture  of  the  plate  along  a 
line  of  rivets.  Ascertain  whether  the  joint  is  proportioned  in 
accordance  with  theory 

Prob.  25.  A  wrought  iron  pipe  f  inches  thick  and  20  inches 
in  diameter  is  to  be  subjected  to  a  head  of  water  of  345  feet. 
Compute  the  probable  increase  in  diameter  due  to  the  interior 
pressure,  regarding  the  pipe  as  thin. 

Prob.  26.  Let  a  pier  whose  top  width  is  b  and  length  /  support 
a  uniformly  distributed  load  P.  Let  the  width  of  the  pier  at  a 
distance  y  below  the  top  be  x,  its  constant  length  being  /  at  all 
horizontal  sections.  Let  w  be  the  weight  of  the  masonry  per 
cubic  foot.  Prove  that,  in  order  to  make  the  compressive 
unit-stress,  the  same  for  all  horizontal  sections,  the  profile  of 

the  pier  must  be  such  as  to  satisfy  the  equation  y  =  --  log, 

w         b 

in  which  .S  =  — -. 
ol 


36  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.        CH.  III. 


CHAPTER   III. 
CANTILEVER  BEAMS  AND  SIMPLE   BEAMS. 

ART.  15.    DEFINITIONS. 

Transverse  stress,  or  flexure,  occurs  when  a  bar  is  in  a 
horizontal  position  upon  one  or  more  supports.  The  weight 
of  the  bar  and  the  loads  upon  it  cause  it  to  bend  and  induce 
in  it  stresses  and  strains  of  a  complex  nature  which,  as  will  be 
seen  later,  may  be  resolved  into  those  of  tension,  compression, 
and  shear.  Such  a  bar  is  called  a  beam. 

A  simple  beam  is  a  bar  resting  upon  supports  at  its  ends. 
A  cantilever  beam  is  a  bar  on  one  support  in  its  middle,  or  the 
portion  of  any  beam  projecting  out  of  a  wall  or  beyond  a  sup* 
port  may  be  called  a  cantilever  beam.  A  continuous  beam  is 
a  bar  resting  upon  more  than  two  supports.  In  this  book  the 
word  beam,  when  used  without  qualification,  includes  all  kinds, 
whatever  be  the  number  of  the  supports,  or  whether  the  ends 
be  free,  supported,  or  fixed. 

The  elastic  curve  is  the  curve  formed  by  a  beam  as  it  deflects 
downward  under  the  action  of  its  own  weight  and  of  the  loads 
upon  it.  Experience  teaches  that  the  amount  of  this  deflection 
and  curvature  is  very  small.  A  beam  is  said  to  be  fixed  at  one 
end  when  it  is  so  arranged  that  the  tangent  to  the  elastic  curve 
at  that  end  always  remains  horizontal.  This  may  be  done  in 
practice  by  firmly  building  one  end  into  a  wall.  A  beam  fixed 
at  one  end  and  unsupported  at  the  other  is  a  cantilever  beam. 

The  loads  on  beams  are  either  uniform  or  concentrated.  A 
uniform  load  embraces  the  weight  of  the  beam  itself  and  any 
load  evenly  spread  over  it.  Uniform  loads  are  estimated  by 
their  intensity  per  unit  of  length  of  the  beam,  and  usually  in 


ART    l6.  REACTIONS   OF   THE   SUPPORTS.  37 

pounds  per  linear  foot.  The  uniform  load  per  linear  unit  is 
designated  by  w,  then  wx  will  represent  the  load  over  any 
distance  x.  If  /  be  the  length  of  the  beam,  the  total  uniform 
load  is  wl  which  may  be  represented  by  W.  A  concentrat- 
ed load  is  a  weight  applied  at  a  definite  ppint  and  is  desig- 
nated by  P. 

In  this  chapter  cantilever  and  simple  beams  will  be  princi- 
pally discussed,  although  all  the  fundamental  principles  and 
methods  hold  good  for  restrained  and  continuous  beams  as 
well.  Unless  otherwise  stated  the  beams  will  be  regarded  as 
of  uniform  cross-section  throughout,  and  in  computing  their 
weights  the  rules  of  Art.  i  will  be  found  of  service. 

Prob.  27.  Find  the  diameter  of  a  round  steel  bar  which 
weighs  48  pounds,  its  length  being  4  feet. 

ART.  16.  REACTIONS  OF  THE  SUPPORTS. 

The  points  upon  which  a  beam  is  supported  react  upward 
against  the  beam  an  amount  equal  to  the  pressure  of  the  beam 
upon  them.  The  beam,  being  at  rest,  is  a  body  in  equilibrium 
under  the  action  of  a  system  of  forces  which  consist  of  the 
downward  loads  and  the  upward  reactions.  The  loads  are 
usually  given  in  intensity  and  position,  and  it  is  required  to 
find  the  reactions.  This  is  effected  by  the  application  of  the 
fundamental  conditions  of  static  equilibrium,  which  for  a 
system  of  vertical  forces,  are, 

2  of  all  vertical  forces  =  o, 

2  of  moments  of  all  forces  =  o. 

The  first  of  these  conditions  says  that  the  sum  of  all  the 
loads  on  the  beam  is  equal  to  the  sum  of  the  reactions.  Hence 
if  there  be  but  one  support,  this  condition  gives  at  once  the 
reaction. 

For  two  supports  the  second  condition  must  be  used  in  con- 
nection with  the  first.  The  center  of  moments  may  be  taken 


38  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

anywhere  in  the  plane,  but  it  is  more  convenient  to  take  it  at  one 
of  the  supports.     For  example,  consider  a  single  concentrated 

load  P  situated  at  4  feet  from  the 
left  end  of  a  simple  beam  whose 
span  is  13  feet.     The  equation  of 
y       moments,  with  the  center   at  the 
**'         left     support,    is     1 3/?a  —  4/>  =  o, 


^ 

from  which  R^  —  ^P.     Again  the 

equation  of  moments,  with  the  center  at  the  right  support,  is 
13^  —  gP=o,  from  which  Rl  =  ^P.  As  a  check  it  may  be 
observed  that  R,  +  R,  =  P. 

For  a  uniform  load  over  a  simple  beam  it  is  evident,  without 
applying  the  conditions  of  equilibrium,  that  each  reaction  is 
one-half  the  load. 

The  reactions  due  to  both  uniform  and  concentrated  loads 
on  a  simple  beam  may  be  obtained  by  adding  together  the 
reactions  due  to  the  uniform  load  and  each  concentrated  load, 
or  they  may  be  computed  in  one  operation.  As  an  example 
of  the  latter  method  let  Fig.  9  represent  a  simple  beam  12  feet 
in  length  between  the  supports  and  weighing  35  pounds  per 
.  linear  foot,  its  total  weight 

I  300  60l  1501  t      •  IT 

,  ,  t  ...          >     J       being     420      pounds.       Let 

H 3—  -4^-2— 4<- 3-  -•*[; 4 — 

A  A'     there  be  three  concentrated 

\E!  -K2|       loads   of   300,  60,   and    150 

Fig-  9-  pounds  placed  at  3,  5,  and 

8  feet  respectively  from  the  left  support.     To  find  the  right 

reaction  R^  the  center  of  moments  is  taken  at  the  left  support, 

and  the  weight  of  the  beam  regarded  as  concentrated  at  its 

middle ;  then  the  equation  of  moments  is, 

Rt  X  12  =  420  X  6+  300  X  3+6o  X  5  +  150  X  8 
from  which  ^  =  410  pounds.     In  like  manner  to  find  R}  the 
center  of  moments  is  taken  at  the  right  support,  and 

Rl  x  12  =  420  x6 


ART.  17.  THE   VERTICAL   SHEAR.  39 

from  which  Rt  =  520  pounds.  As  a  check  the  sum  of  Rl  and 
R9  is  seen  to  be  930  pounds  which  is  the  same  as  the  weight  of 
the  beam  and  the  three  loads. 

When  there  are  more  than  two  supports  the  problem  of  find- 
ing the  reactions  from  the  principles  of  statics  becomes  inde- 
terminate, since  two  conditions  of  equilibrium  are  only  sufficient 
to  determine  two  unknown  quantities.  By  introducing,  how- 
ever, the  elastic  properties  of  the  material,  the  reactions  of 
continuous  beams  may  be  deduced,  as  will  be  explained  in 
Chapter  IV. 

Prob.  28.  A  simple  beam  12  feet  long  weighs  20  pounds  per 
linear  foot  and  carries  a  load  of  500  pounds.  Where  should 
this  load  be  put  so  that  one  reaction  may  be  double  the  other  ? 

Prob.  29.  A  simple  beam  weighing  30  pounds  per  linear  foot 
is  1 8  feet  long.  A  weight  of  700  pounds  is  placed  5  feet  from 
the  left  end  and  one  of  500  pounds  at  8  feet  from  the  right 
end.  Find  the  reactions  due  to  the  total  load. 

ART.  17.    THE  VERTICAL  SHEAR. 

When  a  beam  is  short  it  sometimes  fails  by  shearing  in  a 
vertical  section  as  shown  in  Fig.  10.  The  external  force  which 
produces  this  shearing  on  any  section  i 

is  the  resultant  of  all  the  vertical  forces  | ^J J^ 

on  one  side  of  that  section.  Thus,  in 
the  second  diagram  the  resultant  of  all 
these  external  forces  is  the  loads  and 
the  weight  of  the  part  of  the  beam  on 
the  left  of  the  section  ;  in  the  third  dia- 
gram the  resultant  is  the  loads  and  the 
weight  on  the  right,  or  it  is  reaction  at  pig.  i0. 

the  wall  minus  the  weight  of  the  beam  between  the  wall  and 
the  section. 

'Vertical  Shear '  is  the  name  given  to  the  algebraic  sum  of 
all  external  forces  on  the  left  of  the  section  considered.  Let 


, . 

T2~ 


40  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

it  be  denoted  by  V,  then  for  any  section  of  a  simple  or  can 
tilever,  beam, 

V=  Left  reaction  minus  all  loads  on  left  of  section. 
Here  upward   forces  are   regarded  as  positive  and  downward 
forces  as  negative.      V  is  hence  positive  or  negative  according 
as  the  left  reaction  exceeds  or  is  less  than  the  loads  on  the  left 
of  the  section.     To  illustrate,  consider  a  simple  beam  loaded 

in  any  manner  and  cut  at  any  sec- 

P\                    I      IP  tion  by  a  vertical  plane  mn.     Let 

^ — rh *— * 1  R1  be  the  left  and  R,  the  right  re- 
action. Let  2P,  denote  the  sum 
of  all  the  loads  on  the  left  of  the 


Fig-  "•  section  and  2P9  the  sum  of  those 

on  the  right.     Then,  from  the  definition, 

r=Rl--spl. 

Since  R,  +  R,  =  2P,  +  2P>  it  is  clear  if  R,  —  2P,  =  +  V 
that  Rt  —  2P9  =  —  V,  or  that  the  resultant  of  all  the  external 
forces  on  one  side  of  the  section  is  equal  and  opposite  to  the 
resultant  of  those  on  the  other  side.  They  form,  in  short,  a 
pair  of  shears  acting  on  opposite  sides  of  the  section  and  tend- 
ing to  cause  a  sliding  or  detrusion  along  the  section.  The 
value  of  the  vertical  shear  for  any  section  of  a  simple  beam  or 
cantilever  is  readily  found  by  the  above  equation.  When  R1 
exceeds  2P^ ,  the  vertical  shear  V  is  positive,  and  the  left  part 
of  the  beam  tends  to  slide  upward  relative  to  the  right  part. 
When  R1  is  less  than  2Pt ,  the  vertical  shear  V  is  negative, 
and  the  left  part  tends  to  slide  downward  relative  to  the  other. 
In  the  upper  diagram  of  Fig.  10  the  shear  in  the  left  hand  sec- 
tion is  positive  and  that  in  the  right  hand  section  is  negative. 

The  vertical  shear  varies  greatly  in  value  at  different  sections 
of  a  beam.  Consider  first  a  simple  beam  /  feet  long  and  weigh- 
ing w  pounds  per  linear  foot.  Each  reaction  is  then  \wL 
Pass  a  plane  at  any  distance  x  from  the  left  support,  then  from 


ART.  17.  THE   VERTICAL   SHEAR.  41 

the  definition  the  vertical  shear  for  that  section  is  F—  \wl  —  wx. 
Here  it  is  seen  that  V  has  its  greatest  value  \wl  when  x  =  o, 
that  V  decreases  as  x  increases, 
and  that  V  becomes  o  when  x  —  \L  I  I 
When  x  is  greater  than  £/,  V  is 


negative  and  becomes  —  \wl  when 

x  =  /.  The  equation  F=  ^ee'/  —  wx 

is  indeed  the  equation  of  a  straight  Fig.  ia. 

line,  the  origin  being  at  the  left  support,  and  may  be  plotted 

so  that  the   ordinate  at  any  point  will  represent  the  vertical 

shear   for   the    corresponding  section  of  the  beam,  as  shown 

in  Fig.  12. 

Consider  again  a  simple  beam  as  in  Fig.  13  whose  span  is  12 
feet  and  having  three  loads  of  240,  90,  and  120  pounds,  situated 
3,  4,  and  8  feet  respectively  from 
the  left  support.  By  Art.  16  the 
left  reaction  is  found  to  be  280 
and  the  right  reaction  170  pounds. 
Then  for  any  section  between  the 
left  support  and  the  first  load 
the  vertical  shear  is  F=-|-28o 
pounds,  for  a  section  between  the 
first  and  second  loads  it  is  V =  280  —  240  =  +  40  pounds,  for 
a  section  between  the  second  and  third  loads  V=  280  —  240 
-  90  =  —  50  pounds,  and  for  a  section  between  the  third  load 
and  the  right  support  V  =  280  —  240  —  90  —  120  =  —  170 
pounds,  which  has  the  same  numerical  value  as  the  right  reac- 
tion. By  laying  off  ordinates  upon  a  horizontal  line  a  graphical 
representation  of  the  distribution  of  vertical  shears  throughout 
the  beam  is  obtained. 

For  any  section  of  a  simple  beam  distant  x  from  the  left 
support,  let  Rt  denote  the  left  reaction,  w  the  weight  of  the 
uniform  load  per  linear  unit,  and  2Pt  the  sum  of  all  the  con- 


42  CANTILEVER  BEAMS   AND    SIMPLE   BEAMS.      CH.  IIL 

centrated  loads  between  the  section  and  that  support.     Then 
the  definition  gives, 


as  a  general  expression  for  the  vertical  shear  at  that  section. 

A  cantilever  beam  can  be  so  drawn  that  there  is  no  reaction 
at    the    left    end,  and    for    any    section    V  =  —  wx  —  2P1  . 
•  Thus,   in    Fig.   14,  the  vertical 

shear  for  a  section  in  the 
space  a  is  V  •=•  —  wx,  and  for 
a  section  in  the  space  b  it  is 
V  =  —  wx  —  P,  and  the  graphi- 
cal representation  is  as  shown 
Fig.  i4.  below  the  beam. 

The  vertical  shear  for  any  section  of  a  beam  is  a  measure  of 
the  tendency  to  shearing  along  that  section.  The  above  ex- 
amples show  that  this  is  greatest  near  the  supports.  It  is  rare 
that  beams  actually  fail  in  this  manner,  but  it  is  often  necessary 
to  investigate  the  tendency  to  such  failure. 

Prob^.  30.  A  simple  beam  12  feet  long  and  weighing  20 
pounds  per  linear  foot  has  loads  of  600  and  300  pounds  at  2 
and  4  feet  respectively  from  the  left  end.  Find  the  vertical 
shears  at  several  sections  throughout  the  beam,  and  draw  a 
diagram  to  show  their  distribution. 

ART.  1  8.    THE  BENDING  MOMENT. 

The  usual  method  of  failure  of  beams  is  by  cross-breaking  or 
transverse  rupture.  This  is  caused  by  the  external  forces 
producing  rotation  around  some  point  in  the  section  of  failure, 
Thus,  in  Fig.  14,  let  a  be  the  distance  between  the  end  and  the 
load  P,  and  b  be  the  distance  between  P  and  the  wall.  Then 
the  tendency  of  Pto  cause  rotation  around  a  point  in  the  section 
at  the  wall  is  measured  by  its  moment  Pb  ;  if,  however,  the 


ART.    1  8.  THE   BENDING   MOMENT.  43 

load  were  at  the  end  its  tendency  to  produce  rotation  around 
the  same  point  would  be  measured  by  the  moment  f\a  "+  b). 

1  Bending  moment  '  is  the  name  given  to  the  algebraic  sum 
of  the  moments  of  the  external  forces  on  the  left  of  the  sec- 
tion with  reference  to  a  point  in  that  section.  Let  it  be  de- 
noted by  M.  Then,  for  a  cantilever  or  simple  beam, 

M  •=.  moment  of  reaction  minus  sum  of  moments  of  loads. 

Here  the  moment  of  -upward  forces  is  taken  as  positive  and 
that  of  downward  forces  as  negative.  M  may  hence  be  positive 
or  negative  according  as  the  first  or  second  term  is  the  greater. 

For  a  simple  beam  of  length  /,  uniformly  loaded,  each  reaction 
is  \wl.  For  any  section  distant  x  from  the  left  support  the 


moment  is  M  =  \wl  .  x  —  wx  .  \x,      , 

j^-     —    X"     —  — 

x   being   the  lever  arm   of   the  re-  r-f 

action   Jze'/,  and  \x   the  lever  arm 

of    the   load    wx.       Here   M  =  o 

when  x  =  o  and  also  when  x  =  /, 

and  M  is  a  maximum  when  x  =  \l.  Fis-  »s- 

The  equation,  in  short,  is  that  of  a  parabola  whose  maximum 

ordinate  is  £w/f  and  whose  graphical  representation  is  as  given 

in  Fig.  1  5,  each  ordinate  showing  the  value  of  M  for  the  cor- 

responding value  of  the  abscissa  x. 

Consider  next  a  simple  beam  loaded  with  only  three  weights 
P,  ,  P^  ,  and  P3  .  Here  for  any  section  between  the  left  support  and 
the  first  load  M  =  Rx,  and  for  any  izj  |P2  ,;> 

section  between  the  first  and  second  _t!    11  \ 

loads  M  =  kx—P,  (x—d).    Each  of    ; 
these  expressions  is  the  equation  of 
a  straight  line,  x  being  the  abscissa 
and  J/the  ordinate,  and  the  graphical 

representation  of  bending  moments  is  as  shown  in  Fig.  16.  It 
is  seen  that  for  a  simple  beam  all  the  bending  moments  are 
positive. 


44  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

For  a  cantilever  there  is  no  reaction  at  the  left  end  and  all 
the  bending  moments  are  negative,  the  tendency  to  rotation 
thus  being  opposite  in  direction  to  that  in  a  simple  beam.  For 
instance,  for  a  cantilever  beam  uniformly  loaded  and  having  a 
load  at  the  end  the  bending  moment  is  M  =  —Px—^wx1. 
Here  the  variation  of  moments  may  be  represented  by  a  parab- 
ola, M  being  o  at  the  free  end  and  a  maximum  at  the  wall. 

For  any  given  case  the  bending  moment  at  any  section  may 
be  found  by  using  the  definition  given  above.  The  external 
forces  on  the  left  of  the .  section  are  taken  merely  for  conven- 
ience, for  those  upon  the  right  have  also  the  same  bending 
moment  with  reference  to  the  section.  The  bending  moment 
in  all  cases  is  a  measure  of  the  tendency  of  the  external  forces 
on  either  side  of  the  section  to  turn  the  beam  around  a  point 
in  that  section. 

The  bending  moment  is  a  compound  quantity  resulting  from 
the  multiplication  of  a  force  by  a  distance.  Usually  the  forces 
are  expressed  in  pounds  and  the  distances  in  feet  or  inches  ; 
then  the  bending  moments  are  pound-feet  or  pound-inches. 
Thus  if  a  load  of  500  pounds  be  at  the  middle  of  a  simple 
beam  of  8  feet  span,  the  bending  moment  under  the  load  is, 

M  =  250  X  4  —  i  ooo  pound-feet  =  12  ooo  pound-inches. 
Again  let  a  simple  beam  of  8  feet  span  be  uniformly  loaded 
with  500  pounds  and  have  a  weight  of  200  pounds  at  the  mid- 
dle.    Then  the  bending  moment  at  the  middle  is, 

M  =  350  X  4—250  X  2  =  900  pound-feet. 
Hence  the  tendency  to  rupture  is  less  in  the  second  case  than 
in  the  first. 

Prob.  31.  A  beam  6  feet  long  and  weighing  20  pounds  per 
foot  is  placed  upon  a  single  support  at  its  middle.  Compute  the 
bending  moments  for  sections  distant  I,  2,  3,  4,  and  5  feet  from 
the  left  end,  and  draw  a  curve  to  show  the  distribution  of  mo- 
ments throughout  the  beam. 


ART.  19.   INTERNAL  STRESSES  AND   EXTERNAL   FORCES.          45 

Prob.  32.  A  simple  beam  of  6  feet  span  weighs  20  pounds 
per  linear  foot  and  has  a  load  of  270  pounds  at  2  feet  from  the 
left  end.  Find  the  vertical  shears  for  sections  one  foot  apart 
throughout  the  beam,  and  draw  the  diagram  of  shears.  Find 
the  bending  moments  for  the  same  sections  and  draw  the  dia- 
gram to  represent  them. 

ART.  19.    INTERNAL  STRESSES  AND  EXTERNAL  FORCES. 

The  external  loads  and  reactions  on  a  beam  maintain  their 
equilibrium  by  means  of  internal  stresses  which  are  generated 
in  it.  It  is  required  to  determine  the  relations  between  the  ex- 
ternal forces  and  the  internal  stresses ;  or,  since  the  effect  of 
the  external  forces  upon  any  section  is  represented  by  the  ver- 
tical shear  (Art.  17)  and  by  the  bending  moment  (Art.  18),  the 
problem  is  to  find  the  relation  between  these  quantities  and  the 
internal  stresses  in  that  section. 

Consider  a  beam  of  any  kind  which  is  loaded  in  any  manner. 
Imagine  a  vertical  plane  mn  cutting  the  beam  at  any  cross-sec- 
tion.    In  that  section  there  are  act-  i       |  I     I 
ing  unknown  stresses  of  various  in- 
tensities and    directions.     Let  the 
beam  be  imagined  to  be  separated 
into  two  parts  by  the  cutting  plane 
and  let  forces  X,  Y,  Z,  etc.,  equiv- 


Lu;      i 


•x 

T 


alent  to   the    internal    stresses,  be 

/\  « j 

applied  to  the  section  as  shown  in  ,     i 

Fig.  17.     Then   the  equilibrium  of         x      *jm *-— *- 

each  part  of  the  beam  will  be  undis-  r^x^1 

turbed,  for  each  part  will  be  acted  ~f^ 

upon  by  a  system  of  forces  in  equi- 
librium.    Hence  the  following  fun-  Pig.  I7. 
damental  principle  is  established. 

The  internal  stresses  in  any  cross-section  of  a  beam  hold  irk 
equilibrium  the  external  forces  on  each  side  of  that  section. 


I 


46  CANTILEVER  BEAMS   AND    SIMPLE   BEAMS.      CH.  III. 

This  is  the  most  important  principle  in  the  theory  of  flexure. 
It  applies  to  all  beams,  whether  the  cross-section  be  uniform  or 
variable  and  whatever  be  the  number  of  the  spans  or  the  na- 
ture of  the  loading. 

Thus  in  the  above  figure  the  internal  stresses  X,  Y,  Z,  etc., 
hold  in  equilibrium  the  loads  and  reactions  on  the   left  of  the 
section,  and  also  those  on  the  right.'  Considering  one  part  only 
a  system  of  forces  in   equilibrium   is  seen,  to  which  the  three 
necessary  and  sufficient  conditions  of  statics  apply,  namely, 
2  of  all  horizontal  components  —  o, 
2  of  all  vertical  components  =  o, 
2  of  moments  of  all  forces  =  o. 

From  these  conditions  can  be  deduced  three  laws  concerning 
the  unknown  stresses  in  any  section.  Whatever  be  the  inten- 
sity and  direction  of  these  stresses,  let  each  be  resolved  into 
its  horizontal  and  vertical  components.  The  horizontal  com- 
ponents will  be  applied  at  different  points  in  the  cross-section, 
v  II  some  acting  in  one  direc- 

— 1 — 1 tion  and  some  in  the  other, 

or  in  other  words,  some  of 
the  horizontal    stresses  are 
tensile  and    some  compres- 
Fjg.  18.  sive  ;  by  the  first  condition 

the  algebraic  sum  of  these  is  zero.  The  vertical  components 
will  add  together  and  form  a  resultant  vertical  force  F  which, 
by  the  second  condition,  equals  the  algebraic  sum  of  the  exter- 
nal forces  on  the  left  of  the  section.  As  this  internal  force 
Facts  in  contrary  directions  upon  the  two  parts  into  which  the 
beam  is  supposed  to  be  separated,  it  is  of  the  nature  of  a  shear. 
Hence  for  any  section  of  any  beam  the  following  laws  concern- 
ing the  internal  stresses  may  be  stated. 

1st.  The  algebraic  sum  of  the  horizontal  stresses  is  zero  ;  or 
the  sum  of  the  horizontal  tensile  stresses  is  equal  to  the 
sum  of  the  horizontal  compressive  stresses. 


££ 
I 


ART.   IQ.    INTERNAL   STRESSES   AND    EXTERNAL  FORCES.        47 

2nd.  The  algebraic  sum  of  the  vertical  stresses  forms  a  re- 
sultant shear  which  is  equal  to  the  algebraic  sum  of  the 
external  vertical  forces  on  either  side  of  the  section. 

3rd.  The  algebraic  sum  of  the  moments  of  the  internal 
stresses  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  external  forces  on  either  side  of  the  section. 

These  three  theoretical  laws  are  the  foundation  of  the  theory 
of  the  flexure  of  beams.  Their  expression  may  be  abbreviated 
by  introducing  the  following  definitions. 

4  Resisting  shear*  is  the  name  given  to  the  algebraic  sum  of 
the  internal  vertical  stresses  in  any  section,  and  ' vertical  shear' 
is  the  name  for  the  algebraic  sum  of  the  external  vertical  forces 
on  the  left  of  the  section.  'Resisting  moment'  is  the  name 
given  to  the  algebraic  sum  of  the  moments  of  the  internal  hori- 
zontal stresses  with  reference  to  a  point  in  the  section,  and 
*  bending  moment '  is  the  name  for  the  algebraic  sum  of  the 
moments  of  the  external  forces  on  either  side  of  the  section 
with  reference  to  the  same  point.  Then  the  three  laws  may 
be  thus  expressed  for  any  section  of  any  beam, 

Sum  of  tensile  stresses  =  Sum  of  compressive  stresses. 
Resisting  shear  =  Vertical  shear. 

Resisting  moment          ==  Bending  moment. 

The  second  and  third  of  these  equations  furnish  the  funda- 
mental laws  for  investigating  beams.  They  state  the  relations 
between  the  internal  stresses  in  any  section  and  the  external 
forces  on  either  side  of  that  section.  For  the  sake  of  uniform- 
ity the  external  forces  on  the  left  hand  side  of  the  section  will 
generally  be  used,  as  was  done  in  Arts.  17  and  18. 

Prob.  33.  A  beam  of  weight  W  which  is  6  feet  long  is  sus- 
tained at  one  end  by  a  force  of  280  pounds  acting  at  an  angle 
of  60  degrees  with  the  vertical,  and  at  the  other  end  by  a  ver- 
tical force  Y  and  a  horizontal  force  X.  Find  the  values  of  X 
and  Y,  and  the  weight  of  the  beam. 


48  CANTILEVER  BEAMS  AND  SIMPLE  BEAMS.    CH.  III. 

ART.  20.    EXPERIMENTAL  AND  THEORETICAL  LAWS. 

From  the  three  necessary  conditions  of  static  equilibrium,  as 
stated  in  Art.  19,  three  important  theoretical  laws  regarding 
internal  stresses  were  deduced.  These  alone,  however,  are  not 
sufficient  for  the  full  investigation  of  the  subject,  but  recourse 
must  be  had  to  experience  and  experiment.  Experience  teaches 
that  when  a  beam  deflects  one  side  becomes  concave  and  the 
other  convex,  and  it  is  reasonable  to  suppose  that  the  hori- 
zontal tensile  stresses  are  on  the  convex  side  and  the  compres- 
sive  stresses  on  the  concave.  By  experiments  on  beams  this  is 
confirmed  and  the  following  laws  deduced. 

(F)  —  The  horizontal  fibers  on  the  convex  side  are  elongated 
and  those  on  the  concave  are  shortened,  while  near  the 
center  is  a  '  neutral  surface  '  which  is  unchanged  in  length. 

(G)  —  The  amount  of  elongation  or  compression  of  any  fiber 
is  directly  proportional  to  its  distance  from  the  neutral 
surface.     Hence  by  law  (B)  the  horizontal  stresses  are 
also   directly  proportional  to  their  distances  from  the 
neutral  surface,  provided  the  elastic  limit  of  the  material 
be  not  exceeded.     (See  Art.  50.) 

From  these  laws  there  will  now  be  deduced  the  following  im- 
portant theorem  regarding  the  position  of  the  neutral  surface  : 

The  neutral  surface  passes  through  the  centers  of  gravity 
of  the  cross-sections. 

To  prove  this  let  a  be  the  area  of  any  elementary  fiber  and  z  its 
distance  from  the  neutral  surface.  Let  5  be  the  unit-stress 
on  the  fiber  most  remote  from  the  neutral  surface  at  the 
distance  c.  Then  by  law 


=  unit-stress  at  the  distance  unity, 
c 

—  z  =  unit-stress  at  the  distance  2, 


ART.  21.        THE   TWO   FUNDAMENTAL   FORMULAS.  49 

therefore        -  az  =  the  total  stress  on  any  fiber  of  area  a, 

Sciz 
and  2 =  algebraic  sum  of  all  horizontal  stresses. 

But  by  the  first  law  of  Art.  19  this  algebraic  sum  is  zero,  and 
since  5  and  rare  constants  the  quantity  2azmust  be  zero.  This, 
however,  is  the  condition  which  makes  the  line  of  reference 
pass  through  the  center  of  gravity  as  is  plain  from  the  defini- 
tion of  the  term  'center  of  gravity.'  Therefore,  the  neutral 
surface  of  beams  passes  through 
the  centers  of  gravity  of  the 
cross-sections. 


-!-_,_.,_„        _Neutr.lSurf«. 

The  *  neutral  axis '  of  a  cross- 


section  is  the  line  in  which  the 

neutral    surface    intersects   the 

plane  of  the  cross-section.     On  the  left  of  Fig.  19  is  shown  the 

neutral  axis  of  a  cross-section  and  on  the  right  a  trace  of  the 

neutral  surface. 

Prob.  34.  A  beam  3  inches  wide  and  6  inches  deep  is  loaded 
so  that  the  unit-stress  at  the  remotest  fiber  of  a  certain  cross- 
section  is  600  pounds  per  square  inch.  Find  the  sum  of  all  the 
tensile  stresses  on  the  cross-section. 

Prob.  35.  A  wooden  beam  6  x  12  inches  and  five  feet  long  is 
supported  at  one  end  and  kept  level  by  two  horizontal  forces 
X  and  Z  acting  at  the  other  end  in  the  median  line  of  the  cross- 
section,  the  former  at  2  inches  from  the  top  and  the  latter  at  2 
inches  from  the  base.  Find  the  values  of  X  and  Z. 


ART.  21.  THE  Two  FUNDAMENTAL  FORMULAS. 

Consider  again  any  beam  loaded  in  any  manner  and  cut  at 
any  section  by  a  vertical  plane.  The  internal  stresses  in  that 
section  hold  in  equilibrium  the  external  forces  on  the  left  of 


5O  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.       CK.  HI, 

the  section,  and  as  shown  in  Art.  19,  the  followirg  fundamental 
laws  obtain, 

Resisting  shear        =  Vertical  shear, 
Resisting  moment  =  Bending  moment. 

The  principles  established  in  the  preceding  pages  can  now  be 
applied  to  the  algebraic  expression  of  these  four  quantities. 

The  resisting  shear  is  the  algebraic  sum  of  all  the  vertical 
components  of  the  internal  stresses  at  any  section  of  the  beam. 
If  A  be  the  area  of  that  section  and  Ss  the  shearing  unit-stress, 
regarded  as  uniform  over  the  area,  then  from  formula  (i), 

Resisting  shear  =  ASS. 

The  vertical  shear  for  the  same  section  of  the  beam  being  V 
(Art.  17),  the  first  of  the  above  fundamental  laws  becomes, 

(3)  AS.  =  F, 

which  is  the  first  fundamental  formula  for  the  discussion  of 
beams. 

The  resisting  moment  is  the  algebraic  sum  of  the  moments 
of  the  internal  horizontal  stresses  at  any  section  with  reference 
to  a  point  in  that  section.  To  find  an  expression  for  its  value 
let  5  be  the  horizontal  unit-stress,  tensile  or  compressive  as  the 
case  may  be,  upon  the  fiber  most  remote  from  the  neutral  axis 
and  let  c  be  the  shortest  distance  from  that  fiber  to  said  axis. 
Also  let  z  be  the  distance  from  the  neutral  axis  to  any  fiber 
having  the  elementary  area  a.  Then  by  law  (G)  and  Fig.  19, 

£ 

-  =  unit-stress  at  a  distance  unity, 

£ 

—  z  =  unit-stress  at  distance  z, 


hence  --  =  total  stress  on  any  fiber  of  area  a, 


ART.  21.        THE  TWO   FUNDAMENTAL   FORMULAS.  $1 

and  -  =  moment  of  this  stress  about  neutral  axis. 

c 


Hence      2  --  =  resisting  moment  of  horizontal  stresses. 

Since  5  and  c  are  constants  this  expression  may  be  written 

5 

-2az\     But  2aif,  being  the  sum  of  the  products  formed  by 

multiplying  each  elementary  area  by  the  square  of  its  distance 
from  the  neutral  axis,  is  the  moment  of  inertia  of  the  cross- 
section  with  reference  to  that  axis  and  may  be  denoted  by  /. 

Therefore, 

57 

Resisting  moment  =  -- 

The  bending  moment  for  the  same  section  of  the  beam  being 
M  (Art.  1  8),  the  second  of  the  above  fundamental  laws  becomes, 

(4)  ~  =  M, 

which  is  the  second  fundamental  formula  for  the  discussion  of 
beams. 

Experience  and  experiment  teach  that  simple  beams  of  uni- 
form section  break  near  the  middle  by  the  tearing  or  crushing 
of  the  fibers  and  very  rarely  at  the  supports  by  shearing. 
Hence  it  is  formula  (4)  that  is  mainly  needed  in  the  practical 
investigation  of  beams.  The  following  example  and  problem 
relate  to  formula  (3)  only,  while  formula  (4)  will  receive  detailed 
discussion  in  the  subsequent  articles. 

As  an  example,  consider  a  wrought  iron  I  beam  15  feet  long 
and  weighing  200  pounds  per  yard,  over  which  roll  two  locomo- 
tive wheels  6  feet  apart  and  each  bearing  12  ooo  pounds.  The 
maximum  vertical  shear  at  the  left  support  will  evidently  occur 
when  one  wheel  is  at  the  support  (Art.  16).  The  reaction  will 
then  be  500  -f-  12  coo  +  TV  X  12  ooo  =  19  700  pounds.  Ac- 
cordingly the  greatest  value  of  V  in  the  beam  is  19  700 


52  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

pounds.  As  the  area  of  the  cross-section  is  20  square  inches 
the  average  shearing  unit-stress  by  formula  (3)  is  985  pounds, 
so  that  the  factor  of  safety  is  about  50. 

Prob.  36.  A  wooden  beam  6x9  inches  and  12  feet  in  span 
carries  a  uniform  load  of  20  pounds  per  foot  besides  its  own 
weight  and  also  two  wheels  6  feet  apart,  one  weighing  4  ooo 
pounds  and  the  other  3  ooo  pounds.  Find  the  factor  of  safety 
against  shearing. 

ART.  22.    CENTER  OF  GRAVITY  OF  CROSS-SECTIONS. 

The  fundamental  formula  (4)  contains  c,  the  shortest  dis- 
tance from  the  remotest  part  of  the  cross-section  to  a  hori- 
zontal axis  passing  through  the  center  of  gravity  of  that  cross- 
section.  The  methods  of  rinding  c  are  explained  in  books  on 
theoretical  mechanics  and  will  not  here  be  repeated.  Its  val- 
ues for  some  of  the  simplest  cases  are  however  recorded  for 
reference. 

For  a  rectangle  whose  height  is  d,  c  —  \d. 

For  a  circle  whose  diameter  is  d,  c  =  %d. 

For  a  triangle  whose  altitude  is  d,  c  =  \d 

For  a  square  with  side  d  having  one  diagonal 

vertical,  c  =  d  yf" 

For  a  I  whose  depth  is  d,  c  =  %d. 

For  a  JL  whose  depth  is  d,  thickness  of  flange  £, 
width  of  flange  d,  and  thickness  of  web  t', 


t'd  +  t(b  -  t'} 
For  a  trapezoid  whose  depth  is  d,  upper  base  £, 

b  +  2b'  a 

and  lower  base  b  ,  c  =  -j—.  —  77  •  —  • 

b  +  b     3 

The  student  should  be  prepared  to  readily  apply  the  principle 
of  moments  to  the  deduction  of  the  numerical  value  of  c  for 
any  given  cross-section.  In  nearly  all  cases  the  given  area 
may  be  divided  into  rectangles,  triangles,  and  circular  areas, 


ART.  23.     MOMENT   OF   INERTIA   OF  CROSS-SECTIONS.  53 

whose  centers  of  gravity  are  known,  so  that  the  statement  of 
the  equation  for  finding  c  is  very  simple. 

Prob.  37.  Find  the  value  of  c  for  a  rail  headed  beam  whose 
section  is  made  up  of  a  rectangular  flange  }  X  4  inches,  a  rect- 
angular web  \  X  5  inches,  and  an  elliptical  head  J  inches  deep 
and  i  £  inches  wide. 

ART.  23.    MOMENT  OF  INERTIA  OF  CROSS-SECTIONS. 

The  fundamental  formula  (4)  contains  /,  the  moment  of  in- 
ertia of  the  cross-section  of  the  beam  with  reference  to  a  hori- 
zontal axis  passing  through  the  center  of  gravity  of  that  cross- 
section.  Methods  of  determining  /  are  explained  in  works  on 
elementary  mechanics  and  will  not  here  be  repeated,  but  the 
values  of  some  of  the  most  important  cases  are  recorded  for 
reference. 

r    /» 

For  a  rectangle  of  base  b  and  depth  d,  I  =  -- 

nd% 

For  a  circle  of  diameter  dy  I  =.  —=—  • 

64 

For  an  ellipse  with  axes  a  and  b,  the  latter 

nab* 

vertical,  I——?—  - 

64 

bd* 
For  a  triangle  of  base  b  and  depth  d,  /=  —  • 

For  a  square  with  side  d,  having  one  diag- 

onal vertical,  /=  —  • 

For  a  I  with  base  b,  depth  d,  thickness  of 
flanges  t  and  thickness  of  web  /'  ', 

bd*  —  (b  —  f)(d  —  2t)% 

12 

For  a  JL  with  base  b,  depth  d,  thickness 
of   flange   /,  thickness  of  web  /'  and 


area  A,          /=  -  *  -  k*^,  —  4  —  Ac* 


54  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

The  value  of  /  for  any  given  section  may  always  be  computed 
by  dividing  the  figure  into  parts  whose  moments  of  inertia  are 
known  and  transferring  these  to  the  neutral  axis  by  means  of 
the  familiar  rule  /!  =  /„  +  Ah*,  where  70  is  the  primitive  value 
for  an  axis  through  the  center  of  gravity,  II  the  value  for  any 
parallel  axis,  A  the  area  of  the  figure  and  h  the  distance  be- 
tween the  two  axes. 

Prob.  38.  Compute  the  least  moment  of  inertia  of  a  trape- 
zoid  whose  altitude  is  3  inches,  upper  base  2  inches,  and  lower 
base  5  inches. 

Prob.  39.  Find  the  moment  of  inertia  of  a  triangle  with  ref- 
erence to  its  base,  and  also  with  reference  to  a  parallel  axis 
passing  through  its  vertex. 

ART.  24.    THE  MAXIMUM  BENDING  MOMENT. 

The  fundamental  equation  (4),  namely  —  =  M,  is  true  for 

any  section  of  any  beam,  /  being  the  moment  of  inertia  of  that 
section  about  its  neutral  axis,  c  the  vertical  distance  from  that 
axis  to  the  remotest  fiber,  5  the  tensile  or  compressive  unit- 
stress  on  that  fiber,  and  M  the  bending  moment  of  all  the  ex- 
ternal forces  on  one  side  of  the  section.  For  a  beam  of  con- 
stant cross-section  S  varies  directly  as  M,  and  the  greatest  5 
will  be  found  where  M  is  a  maximum.  The  place  where  M 
has  its  maximum  value  may  hence  be  called  the  *  dangerous 
section,'  it  being  the  section  where  the  horizontal  fibers  are 
most  highly  strained. 

For  a  simple  beam  uniformly  loaded  with  w  pounds  per 
linear  unit,  the  dangerous  section  is  evidently  at  the  middle, 

wl* 
and,  as  shown  in  Art.  18,  the  maximum  Mis  -$-• 

o 

For  a  simple  beam  loaded  with  a  single  weight  P  at  the  dis- 

l  —  p 
tance/  from  the  left  support,  the  left  reaction  is  R  —  P — — , 


ART.  24.  THE   MAXIMUM   BENDING   MOMENT.  55 


—  p\p 

and  the  maximum  moment  is  -  —  ~  —  .  If  Pbe  movable  the 
distance/  will  be  variable,  and  when  the  load  is  at  the  middle 
the  maximum  M  is 


For  a  beam  loaded  with  given  weights,  either  uniform  or 
concentrated,  it  may  be  shown  that  the  dangerous  section  is  at 
the  point  where  the  vertical  shear  passes  through  zero.  To 
prove  this  let  P^  be  any  concentrated  load  on  the  left  of  the 
section  and/  its  distance  from  the  left  support,  and  w  the  uni- 
form load  per  linear  unit.  Then,  for  any  section  distant  x 
from  the  left  support, 

M=Rlx-wx.--  2P,(x  -/). 

To  find  the  value  of  x  which  renders  this  a  maximum,  the  first 
derivative  must  be  put  equal  to  zero  ;  thus, 

~  =  R.-wx  —  2P,  =  o. 
dx 

But  RI  —  wx  —  2P{  is  the  vertical  shear  V  for  the  section  x 
(see  Art.  17).  Therefore  the  maximum  moment  occurs  at  the 
section  where  the  vertical  shear  passes  through  zero. 

To  find  the  dangerous  section  for  any  given  case  the  reac- 
tions are  first  to  be  computed  by  Art.  16,  and  then  the  verti- 
cal shears  are  to  be  investigated  by  Art.  17.  For  a  cantilever, 
however  it  be  loaded,  it  is  seen  that  the  dangerous  section 
is  at  the  wall.  For  a  simple  beam  with  concentrated  loads 
the  point  where  the  vertical  shear  passes  through  zero  must 
usually  be  ascertained  by  trial.  Thus,  referring  to  Fig.  9  and 
the  example  in  Art.  16,  the  vertical  shear  just  at  the  left  of 
the  first  load  is  V  =  520  —  3  X  35  =  +  4r5  pounds,  and  just 
at  the  right  of  the  first  load  it  is  F=  520  —  3X35  —  300  — 
+  115  pounds.  Again  for  the  second  load  the  vertical  shear 
just  at  the  left  is  V=  520  —  5  X  35  —  30x3  =  -f  45  pounds,  and 
just  at  the  right  it  is  V=  520  —  5  X  35  —  360  =  —  15  pounds. 
Hence  in  this  case  the  vertical  shear  changes  sign,  or  passes 


56  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

through  zero,  under  the  second  load,  and  accordingly  this  is 
the  position  of  the  dangerous  section. 

When  the  dangerous  section  has  been  found  the  bending 
moment  for  that  section  is  to  be  computed  by  the  definition 
of  Art.  1  8,  and  this  will  be  the  maximum  bending  moment  for 
the  beam.      Thus,  for  the  numerical  example  of  the  last  para- 
graph, the  maximum  bending  moment  is, 
M=  520  X  5  —  175  X  2j  —  300  X  2  =  +  1562.5  pound-feet. 
Again,  let  a  cantilever  beam  8  feet  long  be  loaded  with  40  pounds 
per  linear  foot  and  carry  a  weight  of  1  50  pounds  at  the  free  end  ; 
then  the  maximum  bending  moment  is, 

M  =  —  320  X  4  —  150x8=—  2  480  pound-feet. 
The  bending  moment  for  simple  beams  is  seen  to  be  always 
positive  and  for  cantilever  beams  always  negative.  That  is  to 
say,  in  the  former  case  the  exterior  forces  on  the  left  of  the 
section  cause  compression  in  the  upper  and  tension  in  the  lower 
fibers  of  the  beam,  while  in  the  latter  case  this  is  reversed  ;  or 
the  upper  side  of  a  deflected  simple  beam  is  concave  and  the 
upper  side  of  a  deflected  cantilever  beam  is  convex. 

Prob.  40.  A  simple  beam  12  feet  long  carries  a  load  of  150 
pounds  at  5  feet  from  the  left  end  and  a  load  of  150  pounds  at 
5  feet  from  the  right  end.  Find  the  dangerous  section,  and  the 
maximum  bending  moment. 

Prob.  41.  A  simple  beam  12  feet  long  weighs  20  pounds  per 
foot  and  carries  a  load  of  100  pounds  at  4  feet  from  the  left  end 
and  a  load  of  50  pounds  at  7  feet  from  tlie  left  end.  Find  the 
dangerous  section,  and  the  maximum  bending  moment. 

ART.  25.    THE  INVESTIGATION  OF  BEAMS. 

The  investigation  of  a  beam  consists  in  deducing  the  greatest 
horizontal  unit-stress  5  in  the  beam  from  the  fundamental  for- 
mula (4).  This  may  be  written, 


<?- 

T' 


ART.  25.  THE    INVESTIGATION   OF  BEAMS.  57 

First,  from  the  given  dimensions  find,  by  Art.  22,  the  value  of  c 
and  by  Art.  23  the  value  of  7.  Then  by  Art.  24  determine  the 
value  of  maximum  M.  From  (4)  the  value  of  S  is  now  known. 
Usually  c  and  7  are  taken  in  inches,  and  M  in  pound-inches ; 
then  the  value  of  5  will  be  in  pounds  per  square  inch. 

The  value  of  vS  will  be  tension  or  compression  according  as 
the  remotest  fiber  lies  on  the  convex  or  concave  side  of  the  beam. 
If  S'  be  the  unit-stress  on  the  opposite  side  of  the  beam  and 
c'  the  distance  from  the  neutral  axis,  then  from  law  (G), 

C         C'  ft 

-  =  ^r  and  S'  =  S-. 

c       c  c 

If  S  be  tension,  S'  will  be  compression,  and  vice  versa.  Some- 
times it  is  necessary  to  compute  S'  as  well  as  S  in  order  to 
thoroughly  investigate  the  stability  of  the  beam.  By  comparing 
the  values  of  S  and  Sf  with  the  proper  working  unit-stresses 
for  the  given  materials  (Art.  8),  the  degree  of  security  of  the 
beam  may  be  inferred. 

As  an  example  consider  a  wrought  iron  I' beam  whose  depth 
is  12  inches,  width  of  flange  4.5  inches,  thickness  of  flange  I  inch 
and  thickness  of  web  0.78  inches.  It  is  supported  at  its  ends 
forming  a  span  of  12  feet,  and  carries  two  loads  each  weighing 
10  ooo  pounds,  one  being  at  the  middle  and  the  other  at  one 
foot  from  the  right  end. 

By  Art.  i,  w  =  56  pounds  per  linear  foot. 

By  Art.  16,  R  =  6169  pounds. 

By  Art.  22,  c  =  6  inches. 

By  Art.  23,  1=  338  inches4. 

By  Art.  24,  x  =  6  feet  for  dangerous  section. 

By  Art.  24,  max.  M  —  36006  X  12  pound-inches. 

Then  from  formula  (4)  the  unit-stress  at  the  dangerous  section  is, 


~      36000  X  12  x  6 

o  =  -  -  -  -  =  7 


700  pounds  per  square  inch. 


58  CANTILEVER   BEAMS   AND    SIMPLE   BEAMS-      CH.  III. 

This  is  the  compressive  unit-stress  on  the  upper  fiber  and  also 
the  tensile  unit-stress  on  the  lower  fiber,  and  being  only  about 
one-third  of  the  elastic  limit  for  wrought  iron  and  about  one- 
seventh  of  the  ultimate  strength  it  appears  that  the  beam  is 
entirely  safe  for  steady  loads  (Art.  8).  It  will  usually  be  best 
in  solving  problems  to  insert  all  the  numerical  values  at  first  in 
the  formula  and  thus  obtain  the  benefit  of  cancellation. 

A  short  beam  heavily  loaded  should  also  be  investigated  for 
the  shearing  stress  at  the  supports  in  the  manner  mentioned  in 
Art.  21,  but  in  ordinary  cases  there  is  little  danger  from  this 
cause.  Thus  for  the  above  example  the  maximum  vertical 
shear  occurs  at  the  right  end  and  is  14500  pounds;  as  the 
area  of  the  cross-section  is  16.8  square  inches,  the  mean  shear- 
ing unit-stress  at  the  right  end  is  from  (3), 

Se  =  —^-jr —  =  863  pounds  per  square  inch, 
ID. 8 

so  that  the  factor  of  safety  against  shearing  is  nearly  60. 

Prob.  42.  A  piece  of  scantling  2  inches  square  and  10  feet 
long  is  hung  horizontally  by  a  rope  at  each  end  and  three 
painters  stand  upon  it.  Is  it  safe? 

Prob.  43.  A  wrought  iron  bar  one  inch  in  diameter  and  two 
feet  long  is  supported  at  its  middle  and  a  load  of  500  pounds 
hung  upon  each  end  of  it.  Find  its  factor  of  safety. 

ART.  26.    SAFE  LOADS  FOR  BEAMS. 

• 

The  proper  load  for  a  beam  should  not  make  the  value  of  5 
at  the  dangerous  section  greater  than  the  allowable  unit-stress. 
This  allowable  unit-stress  or  working  strength  is  to  be  assumed 
according  to  the  circumstances  of  the  case  by  first  selecting  a 
suitable  factor  of  safety  from  Art.  8  and  dividing  the  ultimate 
strength  of  the  material  by  it,  the  least  ultimate  strength  whether 
tensile  or  compressive  being  taken.  For  any  given  beam  the 
quantities  /  and  c  are  known.  Then,  by  the  general  formula  (4), 


ART.  27.  DESIGNING   OF   BEAMS.  59 

the  bending  moment  M  may  be  expressed  in  terms  of  the  un- 
known loads  on  the  beam,  and  thus  those  loads  be  found.  The 
sign  of  the  bending  moment  should  not  be  used  in  (4),  since 
that  sign  merely  denotes  whether  the  upper  fiber  of  the  beam 
is  in  tension  or  compression,  or  indicates  the  direction  in  which 
the  external  forces  tend  to  bend  it. 

As  an  example,  consider  a  cantilever  beam  whose  length  is 
6  feet,  breadth  2  inches,  depth  3  inches  and  which  is  loaded 
uniformly  with  w  pounds  per  linear  foot.  It  is  required  to  find 
the  value  of  w  so  that  S  may  be  800  pounds  per  square  inch. 
Here  c  =  i£  inches,  /=  .g,  and  M  =  36  x  6zv.  Then  from 
formula  (4),  * 

800  X  54 
2i6ze;= — j — ,          whence         w  •=  1 1  pounds. 

Since  a  wooden  beam  2X3  inches  weighs  about  2  pounds  per 
linear  foot,  the  safe  load  in  this  case  will  be  about  9  pounds 
per  foot. 

Prob.  44.  A  wooden  beam  8x9  inches  and  of  14  feet  span 
carries  a  load,  including  its  own  weight,  of  w  pounds  per  linear 
foot.  Find  the  value  of  w  for  a  factor  of  safety  of  10. 

Prob.  45.  A  steel  railroad  rail  of  2  feet  span  carries  a  load 
P  at  the  middle.  If  its  weight  per  yard  is  56  pounds,  /=  12.9 
inches4  and  c  =  2.16  inches,  find  P  so  that  the  greatest  horizon- 
tal unit-stress  at  the  dangerous  section  shall  be  6  ooo  pounds 
per  square  inch. 

ART.  27.    DESIGNING  OF  BEAMS. 

When  a  beam  is  to  be  designed  the  loads  to  which  it  is  to  be 
subjected  are  known,  as  also  is  its  length.  Thus  the  maximum 
bending  moment  may  be  found.  The  allowable  working  strength 
5  is  assumed  in  accordance  with  engineering  practice.  Then 
formula  (4)  may  be  written, 


60  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

and  the  numerical  value  of  the  second  member  be  found.  The 
dimensions  to  be  chosen  for  the  beam  must  give  a  value  of- 

equal  to  this  numerical  value,  and  these  in  general  are  deter- 
mined tentatively,  certain  proportions  being  first  assumed.  The 
selection  of  the  proper  proportions  and  shapes  of  beams  for 
different  cases  requires  much  judgment  and  experience.  But 
whatever  forms  be  selected  they  must  in  each  case  be  such  as 
to  satisfy  the  above  equation. 

For  instance,  a  wrought  iron  beam  of  4  feet  span  is  required 
to  carry  a  rolling  load  of  500  pounds.  Here,  by  Art.  24,  the 
value  of  maximum  M  due  to  the  load  of  500  pounds  is  6  ooo 
pound-inches.  From  Art.  8  the  value  of  5  for  a  variable  load 
is  about  10  ooo  pounds  per  square  inch.  Then, 

/      6000 


c      10  ooo 


=  0.6  inches3. 


An  infinite  number  of  cross-sections  may  be  selected  with  this 
value  of  — .  If  the  beam  is  to  be  round  and  of  diameter  </,  it 

is  known  that  c  =  \d  and  /=  -7— .      Hence, 

itd* 
,  =0.6,         whence  d=  1.83  inches. 

If  the  cross-section  is  to  be  rectangular,  the  dimensions  1x2 
inches  would  give  the  value  of  —  as  f  which  would  be  a  little 

too  large,  but  it  would  be  well  to  use  it  because  the  weight  of 
the  beam  itself  has  not  been  considered  in  the  discussion.  If 
thought  necessary  these  dimensions  may  now  be  investigated 
by  Art.  25  in  order  to  determine  how  closely  the  actual  unit- 
stress  agrees  with  the  value  assumed.  Thus  the  rectangular 
section  I  x  2  inches  weighs  6f  pounds  per  foot ;  the  maximum 


ART.  28.  THE   MODULUS   OF   RUPTURE.  6l 

bending  moment  is  then  6  1 60  pound-inches,  and  the  unit-stress 
is  found  to  be  9  240  pounds  per  square  inch. 

Prob.  46.  Design  a  hollow  circular  wrought  iron  beam  for  a 
span  of  12  feet  to  carry  a  load  of  320  pounds  per  linear  foot. 

Prob.  47.  A  rectangular  wooden  beam  of  14  feet  span  carries 
a  load  of  I  ooo  pounds  at  its  middle.  If  its  width  is  4  inches 
find  its  depth  for  a  factor  of  safety  of  10. 

ART.  28.    THE  MODULUS  OF  RUPTURE. 

The  fundamental  formula  (4)  is  only  true  for  stresses  within 
the  elastic  limit,  since  beyond  that  limit  the  law  (G)  does  not 
hold,  and  .the  horizontal  unit-stresses 
are  no  longer  proportional  to  their  ) 
distances  from  the  neutral  axis,  but 
increase  in  a  less  rapid  ratio.  The 
sketch  shows  a  case  where  the  fiber 


stresses   above  m  and    below  n  have  Flg>  **' 

surpassed  the  elastic  limit.     It  is  however  very  customary  in 
practical  computations  to  apply  (4)  to  the  rupture  of  beams. 

The  '  modulus  of  rupture '  is  the  value  of  5  deduced  from 
formula  (4)  when  the  beam  is  loaded  up  to  the  breaking  point. 
It  is  always  found  by  experiment  that  the  modulus  of  rupture 
does  not  agree  with  either  the  ultimate  tensile  or  compressive, 
strength  of  the  material  but  is  intermediate  between  them.  If 
formula  (4)  were  valid  beyond  the  elastic  limit,  the  value  of  5 
for  rupture  would  agree  with  the  least  ultimate  strength,  with 
tension  in  the  case  of  cast  iron  and  with  compression  in  the  case 
of  timber.  Ductile  materials  like  wrought  iron  and  soft  steel 
have  no  modulus  of  rupture  because  they  continue  to  bend 
indefinitely  under  increasing  transverse  loads  and  failure  does 
not  occur  by  breaking.  The  following  table  gives  values  of 
the  modulus  of  rupture,  all  in  pounds  per  square  inch,  as 
found  by  tests  on  rectangular  beams. 


62 


CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 


Material. 

Tensile 

Strength,  St  . 

Modulus  of 
Rupture,  Sr. 

Compressive 

Strength,  Sc  . 

Timber 

IOOOO 

9000 

8000 

Brick 

800 

2  500 

Stone 

2000 

6  ooo 

Cast  Iron 

20000 

35000 

90000 

Wrought  Iron 

55000 

. 

55000 

Steel 

IOOOOO 

150000 

By  the  use  of  the  experimental  values  off  the  modulus  of  rup- 
ture it  is  easy  with  the  help  of  formula  (4)  to  determine  what 
load  will  cause  the  rupture  of  a  given  beam,  or  what  must  be 
its  length  or  size  in  order  that  it  may  rupture  under  assigned 
loads.  The  formula  when  used  in  this  manner  is  entirely  em- 
pirical and  has  no  rational  basis. 

Prob.  48.  What  must  be  the  size  of  a  square  wooden  beam  of 
8  feet  span  in  order  to  break  under  its  own  weight? 

Prob.  49.  A  cast  iron  cantilever  beam  2  inches  square  and  6 
feet  long  carries  a  load  P  at  the  end.  Find  the  value  of  P  to 
cause  rupture. 

ART.  29.    COMPARATIVE  STRENGTHS. 

The  strength  of  a  beam  is  measured  by  the  load  that  it  can 
carry.  Let  it  be  required  to  determine  the  relative  strength  of 
the  four  following  cases, 

1st,     A  cantilever  loaded  at  the  end  with  W, 
2nd,   A  cantilever  uniformly  loaded  with  W, 
3rd,    A  simple  beam  loaded  at  trie  middle  with  W, 
4th,   A  simple  beam  loaded  uniformly  with  W. 

Let  /  be  the  length  in  each  case.     Then,  from  Art.  24  and  for- 
mula (4), 

For  ist,     M=Wt    and  hence      W  =     -. 


ART.  29.  COMPARATIVE  STRENGTHS.  63 

Wl  SI 

For  2nd,        M=  -  and  hence          W  =  2-,-. 

2  1C 

Wl  SI 

For  3rd,        M  =  -  and  hence          W  =  4-7-. 

4  lc 

Wl  SI 

For  4th,         M •=•  -5-         and  hence          W=  8-7-. 
o  lc 

Therefore  the  comparative  strengths  of  the  four  cases  are  as 
the  numbers  I,  2,  4,  8.  That  is,  if  four  such  beams  be  of  equal 
size  and  length  and  of  the  same  material,  the  2nd  is  twice  as 
strong  as  the  1st,  the  3rd  four  times  as  strong,  and  the  4th  eight 
times  as  strong.  From  these  equations  also  result  the  follow- 
ing important  laws. 

The  strength  of  a  beam  varies  directly  as  5,  directly  as 
7,  inversely  as  c,  and  inversely  as  the  length  /. 

A  load  uniformly  distributed  produces  only  one-half-  as 
much  stress  as  the  same  load  when  concentrated. 

These  apply  to  all  cantilever  and  simple  beams  whatever  be 
the  shape  of  the  cross-section. 

When  the  "cross-section  is  rectangular,  let  b  be.  the  breadth 
and  dfthe  depth,  then  (Art.  23)  the  above  equations  become, 


where  n  is  either  i,  2,  4,  or  8,  as  the  case  may  be.     Therefore, 
The  strength  of  a  rectangular  beam  varies  directly  as  the 
breadth  and  directly  as  the  square  of  the  depth. 

The  reason  why  rectangular  beams  are  put  with  the  greatest 
dimensions  vertical  is  now  apparent. 

To  find  the  strongest  rectangular  beam  that  can  be  cut  from 
a  circular  log  of  given  diameter  D,  it  is  necessary  to  make  bd* 
a  maximum.  Or  the  value  of  b  is  to  be  found  which  makes 
b  (D*  —  F)  a  maximum.  By  placing  the  first  derivative  equal 
to  zero  this  value  of  b  is  readily  found.  Thus, 

b  =  DV  and  d—DV- 


64 


CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.       ClI.  III. 


Fig.  21. 


Hence  very  nearly,  b\  d\ :  5  : 7.  From  this  it  is  evident 
that  the  way  to  lay  off  the  strongest  beam  on 
\  the  end  of  a  circular  log  is  to  divide  the  diameter 
j  into  three  equal  parts,  from  the  points  of  divi- 
J  sion  draw  perpendiculars  to  the  circumference, 
and  then  join  the  points  of  intersection  with  the 
ends  of  the  diameter,  as  shown  in  the  figure. 
The  beam  thus  cut  out  is,  of  course,  not  as  strong  as  the  log, 
and  the  ratio  of  the  strength  of  the  beam  to  that  of  the  log  is 

that  of  their  values  of  -,  which  will  be  found  to  be  about  0.65. 

Prob.  50.  Compare  the  strength  of  a  rectangular  beam  2 
inches  wide  and  4  inches  deep  with  that  of  a  circular  beam  3 
inches  in  diameter. 

Prob.  51.  Compare  the  strength  of  a  wooden  beam  4  X  6 
inches  and  10  feet  span  with  that  of  a  wrought  iron  beam  I  X  2 
inches  and  7  feet  span. 


' 


ART.  30.     IRON  AND  STEEL  I  BEAMS. 

Medium-steel  I  beams  are  rolled  at  present  in  about  thirteen 
different  depths  or  sizes ;  of  each  there  is  a  light  and  a  heavy 
weight,  and  weights  intermediate  in  value 
may  also  be  obtained.  They  are  extensively 
used  in  engineering  and  architecture.  The 
following  table  gives  mean  sizes,  weights, 
and  moments  of  inertia  of  those  steel  beams 
most  commonly  found  in  the  market.  The 
sizes  of  different  manufacturers  agree  as  to 
depth,  but  vary  slightly  with  regard  to  pro- 
portions of  cross-section,  weights  per  foot, 
Fig.  aa.  an(j  mOments  of  inertia.  Fig.  22  shows  the 

proportions  of  the  light  and  heavy  6-inch  beams.    Wrought-iron 
I  beams  were  extensively  used  prior  to  1890,  but  are  now  rarely 


ART.  30. 


IRON  AND  STEEL  I  BEAMS. 


rolled ;  the  table  for  wrought-iron  sections  given  in  previous 
editions  of  this  book  is  hence  here  replaced  by  one  for  steel. 

The  moments  of  inertia  in  the  fourth  column  of  the  table  are 
taken  about  an  axis  perpendicular  to  the  web  at  the  center, 
this  being  the  neutral  axis  of  the  cross-section  when  used  as  a 
beam.  The  values  of  /'  are  with  reference  to  an  axis  coincid- 
ing with  the  center  line  of  the  web  and  are  for  use  in  Chapter  V. 


Depth. 

Weight  per 
Foot. 

Section 
Area. 

Moment  of 
Inertia. 

Section 
Modulus. 

Moment  of 
Inertia. 

Inches. 

Pounds. 

A 
Sq.  Inches. 

Inches*. 

c 
Inches1. 

I* 
Inches*. 

24 

IOO 

29.4 

2380 

198 

48.6 

24 

80 

23.5 

2088 

174 

42.9 

20 

75 

22.1 

I  269 

127 

302 

20 

65 

*I9.I 

I  170 

iif 

27-9 

18 

70 

20.6 

921 

102 

24.6 

18 

55 

15.9 

796 

88.4 

21.2 

15 

55 

15.9 

511 

68.1 

I7.I 

15 

42 

12.5 

442 

58.9 

14.6 

12 

35 

10.3 

228 

38.0 

10.  1 

12 

31* 

9-3 

216 

36.0 

9.50 

10 

40 

u.  8 

159 

31-8 

9.50 

10 

25 

7-4 

122 

24.4 

6.89 

9 

35 

10.3 

112 

24.8 

7-31 

9 

21 

6.3 

85.0 

18.9 

5-16 

8 

25* 

7-50 

68.4 

17.1 

4-75 

8 

18 

5-33 

56.9 

14.2 

3.78 

7 

20 

5-88 

42.2 

12.  1 

3-24 

7 

15 

4.42 

36.2 

10.4 

2.67 

6 

17^ 

5-07 

26.2 

8-73 

2.36 

6 

I2i 

3-61 

21.8 

7.27 

1.85 

5 

I4f 

4-34 

15.2 

6.08 

1.70 

5 

ti 

2.87 

12.  1 

4.84 

1.23 

4 

xoi 

3.09 

7-1 

3-55 

1.  01 

4 

7* 

2.21 

6.0 

3.00 

0.77 

3 

74 

2.21 

2.9 

1-93 

0.60 

3 

5i 

1.63 

2.5 

1.71 

0.46 

66  CANTILEVER    BEAMS   AND    SIMPLE   BEAMS.      CH.  III. 

In  investigating  the  strength  of  a  given  I  beam  the  value  of  - 

is  taken  from  the  table  and  5  is  computed  from  formula  (4). 
In  designing  an  I  beam  for  a  given  span  and  loads  the  value  of 

-  is  found  by  (4)  from  the  data  and  then  from  the  table  that  I 

is  selected  which  has  the  nearest  or  next  highest  corresponding 
value.     Intermediate  weights  between  those  given  in  the  table 

can  also  usually  be  obtained ;  thus,  if  the  computed  value  of  - 

should  be  37.0  a  12-inch  beam  weighing  about  33  pounds  per 
foot  might  be  chosen. 

For  example,  let  it"  be  required  to  determine  which  I  should 
be  selected  for  a  floor  loaded  with  1 50  pounds  per  square  foot, 
the  beams  to  be  of  20  feet  span  and  spaced  12  feet  apart  be- 
tween centers,  and  the  maximum  unit-stress  5  to  be  16000 
pounds  per  square  inch.  Here  the  uniform  load  on  the  beam 
is  12  X  20  X  150  =  36  ooo  pounds  =  W.  From  formula  (4), 

-  —  HL  —  36000  X  20  X   12  _  ~ 

c  ~~~-  S  "         8X16000  °7'5' 

and  hence  the  heavy  1 5-inch  I  should  be  selected. 

In  the  following  examples  and  problems  the  ultimate  tensile 
strength  of  medium  steel  is  taken  as  65  ooo  pounds  per  square 
inch,  and  factors  of  safety  are  based  on  this  constant. 

Prob.  52.  A  heavy  15  inch  I  beam  of  12  feet  span  sustains  a 
uniformly  distributed  load  of  41  net  tons.  Find  its  factor  of 
safety.  Also  the  factor  of  safety  for  a  24  feet  span  under  the 
same  load. 

Prob.  53.  A  floor,  which  is  to  sustain  a  uniform  load  of  175 
pounds  per  square  foot,  is  to  be  supported  by  heavy  10  inch  I 
beams  of  15  feet  span.  Find  their  proper  distance  apart  from 
center  to  center  so  that  the  maximum  fiber  stress  may  be 
1 2  ooo  pounds  per  square  inch. 


ART.  31. 


IRON   AND   STEEL   DECK   BEAMS. 


Q 


ART.  31.     IRON  AND  STEEL  DECK  BEAMS. 
Deck  beams  are  used  in  the  construction  of  buildings,  and 
are  of  a  section  such  as  shown  in  Fig.  23.     The  heads  are 
formed  with   arcs  of   circles   but    may  be 
taken  as  elliptical  in  computing  the  values 
of  c  and  /.      The  following  table  gives  di- 
mensions   of   a  few  of  the  steel  sections 
found  in  the  market. 

By  means  of  formula  (4)  a  given  deck 
beam  may  be  investigated  or  safe  loads  be 
determined  for  it,  or  one  may  be  selected 
for  a  given  load  and  span.  Sometimes  T 
irons  are  used  instead  of  deck  beams ;  the 
values  of  c  and  /  for  these  are  given  in  the 
handbooks  issued  by  the  manufacturers,  or  they  may  be  com- 
puted  with  an  accuracy  usually  sufficient  by  regarding  the 
web  and  flange  as  rectangular  (Arts.  22  and  23). 


Fig.  23. 


Size.    Depth. 

Weight 
per  Foot. 

Section 
Area. 

Moment 
of  Inertia. 

From  Top 
to  Axis. 

~i 
Section 

Modulus. 
/ 

A 

/ 

e 

c 

Inches. 

Pounds. 

Sq.  Inches. 

Inches*. 

Inches. 

Inches*. 

Heavy  9 

30 

8.8 

93-2 

4-75 

19.6 

Light    9 

26 

7.6 

85.2 

4.81 

17-7 

H           8 

24.5 

7-2 

62.8 

4-45 

I4.I 

L           8 

2O.  I 

5-9 

55-6 

4-56 

12.2 

H           7 

23-5 

6.9 

45-5 

3-89 

II.7 

L           7 

18.1 

5-3 

38.8 

4.00 

9-7 

Prob.  54.  A  heavy  7  inch  deck  beam  is  loaded  uniformly 
with  50  ooo  pounds.  Find  its  factor  of  safety  for  a  span  of 
22  feet.  Also  for  a  span  of  1 1  feet. 

Prob.  55.  What  uniform  load  should  be  placed  upon  a  heavy 
7  inch  deck  beam  of  22  feet  span  so  that  the  greatest  unit-stress 
at  the  dangerous  section  may  be  12  ooo  pounds  per  square  inch? 


68  CANTILEVER  BEAMS  AND  SIMPLE  BEAMS.    CH.  III. 

ART.  32.    CAST  IRON  BEAMS. 

Wrought  iron  beams  are  usually  made  with  equal  flanges 
since  the  resistance  of  wrought  iron  is  about  the  same  for  both 
tension  and  compression.  For  cast  iron,  however,  the  flange 
under  tension  should  be  larger  than  that  under  compression, 
since  the  tensile  resistance  of  the  material  is  much  less  than  its 
compressive  resistance.  Let  S'  be  the  unit-stress  on  the  re- 
motest fiber  on  the  tensile  side  and  S  that  on  the  compressive 
side,  at  the  distances  c'  and  c  respectively  from  the  neutral  axis. 
Then,  from  law 


c'  ~  S'  ' 

Now  if  the  working  values  of  5  and  S'  can  be  selected  the 
ratio  of  c  to  c'  is  known  and  a  cross-section  can  be  designed, 
but  it  is  difficult  to  assign  these  proper  values  on  account  of 
our  lack  of  knowledge  regarding  the  elastic  limits  of  cast  iron. 

According  to   HODGKINSON'S  investigations   the   following 
are  dimensions  For  a  cast  iron  beam  of  equal  ultimate  strength. 

Thickness  of  web  =         /, 

Depth  of  beam  =  13.5^, 

Width  of  tensile  flange  =     \2t, 

Thickness  of  tensile  flange  =       2/, 

Width  of  compressive  flange  =       5/, 
Thickness  of  compressive  flange     =     i%ty 

Value  of  c  =       gt, 

Value  of  /  =  923^. 

Here  the  unit-stress  in  the  tensile  flange  is  one-half  that  in  the 
compressive  flange.  Although  these  proportions  may  be  such 
as  to  allow  the  simultaneous  rupture  of  the  flanges,  yet  it  does 
not  necessarily  follow  that  they  are  the  best  proportions  for 
ordinary  working  stresses,  since  the  factors  of  safety  in  the 
flanges  as  computed  by  the  use  of  formula  (4)  would  be  quite 
different.  The  proper  relative  proportions  of  the  flanges  of 


ART.  32.  CAST  IRON   BEAMS.  69 

cast  iron  beams  for  safe  working  stresses  have  never  been 
definitely  established,  and  on  account  of  the  extensive  use  of 
wrought  iron  the  question  is  not  now  so  important  as  formerly. 

As  an  illustration  of  the  application  of  formula  (4)  let  it  be 
required  to  determine  the  total  uniform  load  W  for  a  cast  iron 
JL  beam  of  14  feet  span,  so  that  the  factor  of  safety  may  be  6, 
the  depth  of  the  beam  being  18  inches,  the  width  of  the  flange 
12  inches,  the  thickness  of  the  stem  I  inch,  and  the  thickness 
of  the  flange  i£  inches.  First,  from  Art.  22  the  value  of  c  is 
found  to  be  12.63  inches,  and  that  of  c'  to  be  5.37  inches, 
From  Art.  23  the  value  of  /  is  computed  to  be  I  031  inches4, 
From  Art.  24  the  maximum  bending  moment  is, 

wl* 

M  =  —  -  =  2  1  W  pound-inches. 
8 

Now  with  a  factor  of  safety  of  6  the  working  strength  S  on  the 
remotest  fiber  of  the  stem  of  the  dangerous  section  is  to  be 

^—7:  —  pounds  per  square  inch.     Hence  from  formula  (4), 


2lW  =  9QQQQX  103^      whence       w  _  53  300  pounds. 

Again  with  a  factor  of  safety  of  6  the  working  strength  S'  on 
the  remotest  fiber  of  the  flange  at  the  dangerous  section  is  to  be 

—  -  —  pounds  per  square  inch.     Hence  from  the  formula, 
2  1  w  =  2Q^°OX  I031,      whence       W  =  30  400  pounds. 

The  total  uniform  load  on  the  beam  should  hence  not  exceed 
30400  pounds.  Under  this  load  the  factor  of  safety  on  the 
tensile  side  is  6,  while  on  the  compressive  side  it  is  nearly  12. 

Prob.  56.  A  cast  iron  beam  in  the  form  of  a  channel,  or 
hollow  half  rectangle,  is  often  used  in  buildings.  Suppose  the 
thickness  to  be  uniformly  one  inch,  the  base  8  inches,  the  height 


70  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

6  inches  and  the  span  12  feet.     Find  the  values  of  5  and  £'  at 
the  dangerous  section  under  a  uniform  load  of  5  ooo  pounds. 

ART.  33.    GENERAL  EQUATION  OF  THE  ELASTIC  CURVE. 

When  a  beam  bends  under  the  action  of  exterior  forces  the 
curve  assumed  by  its  neutral  surface  is  called  the  elastic  curve. 
It  is  required  to  deduce  a  general  expression  for  its  equation. 

Let  //  in  the  figure  be  any  normal  section  in  any  beam. 
Let  mn  be  any  short  length  dl,  measured  on  the  neutral  surface, 

and  let  qmq  be  drawn  parallel  to 
the  normal  .section  through  n. 
Previous  to  the  bending  the  sec- 
tions pp  and  tt  were  parallel ; 
now  they  intorsect  at  o  the  cen- 
ter of  curvature.  Previous  to 
the  bending  pt  was  equal  to  dl, 
now  it  has  elongated  or  shortened 
the  amount  pq.  The  distance 
Fig- 34>  pq  will  be  called  A  and  the  dis- 

tance mp  is  the  quantity  c  (Art.  22).     The  elongation  A  is  pro- 
duced by  the  unit-stress  5,  and  from  (2)  its  value  is, 

A  =  — 

where  E  is  the  coefficient  of  elasticity  of  the  material  of  the 
beam.     From  the  similar  figures  omn  and  mpq, 

om  _  mp  R  __  c 

where  R  is  the  radius  of  curvature  om.     Inserting  in  this  the 
above  value  of  A,  it  becomes, 

S__EL 

f~   R  ' 


ART.  33.    GENERAL   EQUATION   OF  THE   ELASTIC   CURVE.       /I 

But  the  fundamental  formula  (4)  may  be  written  in  the  form 

S_M 

7~=7* 

and  hence,  by  comparison, 

M      EI 
M=—. 

This  is  the  formula  which  gives  the  relation  between  the  bend- 
ing moment  of  the  exterior  forces  and  the  radius  of  curvature 
at  any  section.  Where  M  is  o  the  radius  R  is  oo  ;  where  M  is 
a  maximum  R  has  its  least  value. 

Now,  in  works  on  the  differential  calculus,  the  following 
value  is  deduced  for  the  radius  of  curvature  of  any  plane  curve 
whose  abscissa  is  x,  ordinate  y,  and  length  /,  namely, 


~ 


dx  .  d*y 
Hence  the  most  general  equation  of  the  elastic  curve  is, 

dr       =  £I 

dx  .  d*y  ~  M  ' 

which  applies  to  the  flexure  of  all  bodies  governed  by  the  laws 
of  Arts.  3  and  20. 

In  discussing  a  beam  the  axis  of  x  is  taken  as  horizontal  and 
that  of  y  as  vertical.  Experience  teaches  us  that  the  length  of 
a  small  part  of  a  bent  beam  does  not  materially  differ  from 
that  of  its  horizontal  projection.  Hence  dl  may  be  placed 
equal  to  dx  for  all  beams,  and  the  above  equation  reduces  to 
the  form, 

(5)  ^  =  - 

dx*      EI' 

This  is  the  general  equation  of  the  elastic  curve,  applicable  to 
all  beams  whatever  be  their  shapes,  loads  or  number  of  spans. 
M  is  the  bending  moment  of  the  external  forces  for  any  sec- 


72  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.       CH.  III. 

tion  whose  abscissa  is  x,  and  whose  moment  of  inertia  with 
respect  to  the  neutral  axis  is  /.  Unless  otherwise  stated  /will 
be  regarded  as  constant,  that  is,  the  cross-section  of  the  beam 
is  constant  throughout  its  length. 

To  obtain  the  particular  equation  of  the  elastic  curve  for  any 
special  case,  it  is  first  necessary  to  express  M  as  a  function  of 
x  and  then  integrate  the  general  equation  twice.  The  ordinate 
y  will  then  be  known  for  any  value  of  x.  It  should,  however, 
be  borne  in  mind  that  formula  (5),  like  formula  (4),  is  only  true 
when  the  unit-stress  5  is  less  than  the  elastic  limit  of  the 
material. 

Prob.  57.  A  wooden  beam  £  inch  wide,  f  inch  deep,  and  3 
feet  span  carries  a  load  of  14  pounds  at  the  middle.  Find  the 
radius  of  curvature  for  the  middle,  quarter  points,  and  ends. 

ART.  34.    DEFLECTION  OF  CANTILEVER  BEAMS. 
Case  I.  A  load  at  the  free  end.  —  Take  the  origin  of  co-ordi- 


^ I Jp    ...      [  t _ nates  at  the  free  end,  and  as 

p 


=7  I   I/   in  Fig.  25,  let  m  be  any  point 

^^^_^_  __$  of    the    elastic    curve    whose 

Fig- 25-  abscissa  is  x  and  ordinate  jr. 

For  this  point  the  bending  moment  M  is  —  Px  and  the  general 
formula  (5)  becomes, 


. 

dx* 
By  integration  the  first  derivative  is  found  to  be 

Eg-=-**+C. 

dx  2 

d  * 

But  ~-  is  the  tangent  of  the  angle  which  the  tangent  to  the 
dx 

elastic  curve  at  m  makes  with  the  axis  of  x,  and  as  the  beam  is 


ART.  34.        DEFLECTION   OF   CANTILEVER   BEAMS.  73 

dy 

fixed  at  the  wall  the  value  of  -y-  is  o  when  x  equals  /.     Hence 
C  =  J/Y2,  and  the  first  differential  equation  is, 


'dx~       2  2 

The  second  integration  now  gives, 


But  y  =  o,  when  x  =  o.     Hence  C'  =  o,  and 
6EIy  =  P($rx  -  **), 

which  is  the  equation  of  the  elastic  curve  for  a  cantilever  of 
length  /  with  a  load  P  at  the  free  end.  If  x  =  I  the  value  of 
y  will  be  the  maximum  deflection,  which  may  be  represented 
by  A.  Then, 


and  for  any  point  of  the  beam  the  deflection  is  A  —  y. 

Case  II.  A  uniform  load.  —  Let  the  origin  be  taken  at  the 
free  end  as  before,  and   x  and  y 
be  the  co-ordinates  of  any  point 
of  the  elastic  curve.     Let  the  load 
per  linear  unit  be  w.     Then   for  Fi£- 

any  section  M  =  —  \wx*  and  formula  (5)  becomes, 

wx* 


Integrate  this,  determine  the  constant  of  integration  by  the 

dy 
consideration  tjiat  -r-  =  o  when  x  =  /,  and  then, 

~-=wr  -wx\ 
dx 


74  CANTILEVER  BEAMS  AND    SIMPLE   BEAMS.       CH.  III. 

Integrate  again,  and  after  determining  the  constant,  the  equa- 
tion of  the  elastic  curve  is, 


which  is  a  biquadratic  parabola.     For  x  =  /,  y  =  A  the  maxi 
mum  deflection,  whose  value  is, 


where  W  is  the  total  uniform  load  on  the  cantilever. 

Case  III.  A  load  at  the  free  end  and  also  a  uniform  load.  — 
Here  it  is  easy  to  show  that  the  maximum  deflection  is 

A  _  spr  +  3  wr 

24£7 

which  is  the  sum  of  the  deflections  due  to  the  two  loads. 
Hence  it  appears  that,  as  in  cases  of  stress,  each  load  produces 
its  effect  independently  of  the  other. 

In  order  that  the  formulas  for  deflection  may  be  true,  the 
maximum  unit-stress  5  produced  by  all  the  loads  must  not 
exceed  the  elastic  limit  of  the  material. 

Prob.  58.  Compute  the  deflection  of  a  cast  iron  cantilever 
beam,  2X2  inches  and  6  feet  span,  caused  by  a  load  of  100 
pounds  at  the  end. 

Prob.  59.  In  order  to  find  the  coefficient  of  elasticity  of  a 
cast  iron  bar  2  inches  wide,  4  inches  deep,  and  6  feet  long,  it 
was  balanced  upon  a  support  and  a  weight  of  4  ooo  pounds 
hung  at  each  end,  causing  a  deflection  of  0.401  inches.  Com- 
pute the  value  of  E. 

ART.  35.    DEFLECTION  OF  SIMPLE  BEAMS. 

The  deflection  of  a  simple  beam  due  to  a  load  at  the  middle, 
or  to  a  uniform  load,  is  readily  obtained  from  the  expressions 
just  deduced  for  cantilever  beams.  Thus,  for  a  simple  beam 
of  span  /  with  a  load  P  at  the  middle,  if  Fig.  27  be  inverted  it 


ART.  35.  DEFLECTION   OF   SIMPLE   BEAMS.  75 

will  be  seen  to  be  equivalent  to  two  cantilever  beams  of  length 
\l  with  a  load  \P  at  each  end.  The  formula  for  the  maximum 
deflection  of  a  cantilever  beam  hence  applies  to  this  figure,  if 

pr 

I  be  replaced  by  £/  and  P  by  \P,  which  gives  J  =       ,,  -  for  the 

deflection  at  the  middle  of  the  simple  beam.  It  will  be  well, 
however,  to  use  the  general  formula  (5)  and  treat  each  case  in- 
dependently. 

Case  I.  A  single  load  P  at  the  middle. — Let  the  origin  be 
taken  at  the  left  support.     For 
any   section    between    the    left 
support     and    the    middle    the 
bending    moment    M    is    \Px.  Fie- 

Then  the  general  formula  (5)  becomes, 


dy 
Integrate  this  and  find  the  constant  by  the  fact  that  -7-  =  o 

when  x  =  J/.    Then  integrate  again  and  find  the  constant  by 
the  fact  that  7  =  0  when  x  =  o.     Thus, 

4&EIy  =  P(AfX*  —  $l*x), 

is  the  equation  of  elastic  curve  between  the  left  hand  support 
and  the  load.     For  the  greatest  deflection  make  x  =  J/,  then, 

z>/8 
A  = 


Case  II.  A  uniform  load. — Let  w  be  the  load  per  linear  unit, 
then  the  formula  (5)  becomes, 

d^y  _  wlx      wx* 

hld?  =  T     ~^~- 

Integrate  this  twice,  find  the  constants  as  in  the  preceding  para- 
graph, and  the  equation  of  the  elastic  curve  is, 


76  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

from  which  the  maximum  deflection  is  found  to  be, 

=  384^7  ==:  384^7* 

Case  III.  A  load  P  at  any  point. — Here  it  is  necessary  first 
to  consider  that  there  are  two  elastic  curves,  one  on  each  side 

of  the  load,  which  have  distinct 
equations,  but  which  have  a 
common  tangent  and  ordinate 
under  the  load.  As  in  Fig.  28, 

let  the  load  be  placed  at  a  distance  kl  from  the  left  support, 
k  being  a  number  less  than  unity.     Then  the  left  reaction  is 
R  =  P(\  —  k).     From  the  general  formula  (5),  with  the  origin 
at  the  left  support,  the  equations  are, 
On  the  left  of  the  load, 


(a)        EI        = 


(c) 
On  the  right  of  the  load, 

(a)'       EId-       =  Rx~P(x-  kl\ 


(b)r        EI       =  \Rx*  -  \Px*  +  Pklx  +  C3  , 

(c)'  Ely  =  %Rx*  -  \Px*  +  \Pklx*  +  C,x  +  Ct  . 

To  determine  the  constants  consider  in  (c)  that  y  =  o  when 
x  —  o,  and  hence  that  £7,  =  o.  Also  in  (c)',  y  =  o  when  x  —  l\ 
again  since  the  curves  have  a  common  tangent  under  the  load, 
(b)  =  (b)f  when  x  =  £/,  and  since  they  have  a  common  ordinate 
at  that  point  (c)  =  (c)f  when  x  —  kl.  Or, 

o  = 


ART.  36.    COMPARATIVE   DEFLECTION  AND   STIFFNESS.  // 

From  these  three  equations  the  values  of  Clt  Cit  and  C^  are 
found.  Then  the  equation  of  the  elastic  curve  on  the  left  of 
the  load  is, 

6EIy  =  P(i  -  k)x*  -  P(2k  -  3/P 


To  find  the  maximum  deflection,  the  value  of  x  which  renders 
y  a  maximum  is  to  be  obtained  by  equating  the  first  derivative 
to  zero.  If  k  be  greater  than  £,  this  value  of  x  inserted  in  the 
above  equation  gives  the  maximum  deflection  ;  if  k  be  less  than 
£,  the  maximum  deflection  is  on  the  other  side  of  the  load. 
For  instance,  if  k  =  £,  the  equation  of  the  elastic  curve  on  the 
left  of  the  load  is, 

=  i6/V  - 


This  is  a  maximum  when  x  =  0.567,  which  is  the  point  of  great- 
est deflection. 

Prob.  60.  Prove,  when  k  is  greater  than  \  in  Fig.  28,  that  the 

pr 

maximum  deflection  is  A  =  —=r-r(i  —  k)($k  — 


Prob.  61.  In  order  to  find  the  coefficient  of  elasticity  of 
Quercus  alba  a  bar  4  centimeters  square  and  one  meter  long 
was  supported  at  the  ends  and  loaded  in  the  middle  with  weights 
of  50  and  100  kilograms  when  the  deflections  were  found  to  be 
6.6  and  13.0  millimeters.  Show  that  the  mean  value-  of  E  was 
74  500  kilos  per  square  centimeter. 


ART.  36.    COMPARATIVE  DEFLECTION  AND  STIFFNESS. 

From  the  two  preceding  articles  the  following  values  of  the 
maximum  deflections  may  now  be  written  and  their  compari- 
son will  show  the  relative  stiffness  of  the  different  cases. 

i    wr 

For  a  cantilever  loaded  at  the  end  with  W,         A  =      -  .  — — . 


78  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      Cll.  III. 

i    wr 

For  a  cantilever  uniformly  loaded  with  W,          A  =     -  .  -=rjr. 

8     El 

i     wr 

For  a  simple  beam  loaded  at  middle  with  W,      A  =   —  .  ~—  —  . 

48       El 

t        J^/73 

For  a  simple  beam  uniformly  loaded  with  Wy     A  =  —  1~  .  _  —  . 

384     hi 

The  relative  deflections  of  these  four  cases  are  hence  as  the 
numbers  I,  f  ,  TV,  and 


These  equations  also  show  that  the  deflections  vary  directly 
as  the  load,  directly  as  the  cube  of  the  length,  and  inversely  as 

E  and  /.     For  a  rectangular  beam  I  =  —  —  ,  and  hence  the  de- 

flection of  a  rectangular  beam  is  inversely  as  its  breadth  and 
inversely  as  the  cube  of  its  depth. 

The  stiffness  of  a  beam  is  indicated  by  the  load  that  it  can 
carry  with  a  given  deflection.  From  the  above  it  is  seen  that 
the  value  of  the  load  is, 


where  m  has  the  value  3,  8,  48,  or  ^-f^  as  the  case  may  be. 
Therefore,  the  stiffness  of  a  beam  varies  directly  as  E,  directly 
as  7,  and  inversely  as  the  cube  of  its  length,  and  the  relative 
stiffness  of  the  above  four  cases  is  as  the  numbers  i,  2f,  16, 
and  25f  .  From  this  it  appears  that  the  laws  of  stiffness  are 
very  different  from  those  of  strength.  (Art.  29.) 

Prob.  62.  Compare  the  strength  and  stiffness  of  a  joist  3X8 
inches  when  laid  with  flat  side  vertical  and  when  laid  with  nar- 
row side  vertical. 

Prob.  63.  Find  the  thickness  of  a  white  pine  plank  of  8  feet 
span  required  not  to  bend  more  than  ^f^th  of  its  length  under 
a  head  of  water  of  20  feet. 


ART.  37.    RELATION   BETWEEN  DEFLECTION   AND   STRESS.     79 


ART.  37.    RELATION  BETWEEN  DEFLECTION  AND  STRESS. 

Let  the  four  cases  discussed  in  Arts.  29  and  36  be  again  con- 
sidered.    For  the  strength, 


W  =  n-—,     where  n  =  I,  2,  4,  or  8. 


For  the  stiffness, 


EIA 


W=m  — — ,     where  m  =  3,  8,  48,  or  76^. 

By  equating  these  values  of  W  the  relation  between  A  and  5 
is  obtained,  thus, 


5  = 


mEcA 


or 


A  = 


nl*S 


mcE  ' 

These  equations,  like  the  general  formula  (4)  and  (5),  are  only 
valid  when  5  is  less  than  the  elastic  limit  of  the  materials. 

This  also  shows  that  the  maximum  deflection  A  varies  as 

-  for  beams  of  the  same  material  under  the  same  unit-stress  S. 
c 

From  the  preceding  articles  the  following  table  may  also  be 
compiled  which  exhibits  the  most  important  results  relating  to 
both  absolute  and  relative  strength  and  stiffness. 


Case. 

Max. 
Vertical 
Shear. 

Max. 
Bending 
Moment. 

Max. 
Stress 
S. 

Max. 
Deflec- 
tion. 

Relative 
Strength. 

Relative 
Stiffness. 

Wlc 

I  Wl* 

Cantilever  loaded  at  end, 

W 

Wl 

I 

I 

I 

3  El 

Cantilever  loaded  uniformly, 

W 

km 

Wlc 

7.1 

I  Wl* 

2 

n 

Simple  beam  loaded  at  middle, 

itr 

1    M/7 

Wlc 

I    Wl* 

4 

16 

Simple  beam  loaded  uniformly, 

** 

1    M/7 

Wlc 

5     Wl* 

8 

25* 

80  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

Here  the  signs  of  the  maximum  shears  and  moments  are 
omitted  as  only  their  absolute  values  are  needed  in  computa- 
tions. Evidently  the  moments  are  negative  for  the  first  and 
second  cases,  and  positive  for  the  third  and  fourth,  the  direc- 
tion of  the  curvature  being  different. 

Prob.  64.  Find  the  deflection  of  a  medium-steel  heavy  10- 
inch  I  beam  of  9  feet  span  when  stressed  by  a  uniform  load 
up  to  30000  pounds  per  square  inch. 

Prob.  65.  A  wooden  beam  of  breadth  &,  depth  d,  and  span  x 
is  loaded  with  P  at  the  middle.  Find  the  value  of  x  so  that 
rupture  may  occur  under  the  load.  Find  also  the  value  of  x 
so  that  rupture  may  occur  by  shearing  at  the  supports. 

ART.  38.    CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH. 

All  cases  thus  far  discussed  have  been  of  constant  cross- 
section  throughout  their  entire  length.  But  in  the  general 
formula  (4)  the  unit-stress  5  is  proportional  to  the  bending  mo- 
ment My  and  hence  varies  throughout  the  beam  in  the  same 
way  as  the  moments  vary.  Hence  some  parts  of  the  beam  are 
but  slightly  strained  in  comparison  with  the  dangerous  section. 

A  beam  of  uniform  strength  is  one  so  shaped  that  the  unit- 
stress  S  is  the  same  in  all  fibers  at  the  upper  and  lower  surfaces. 
Hence  to  ascertain  the  form  of  such  a  beam  the  unit-stress  S 

in  (4)  must  be  taken  as  constant  and  —  be  made  to  vary  with 

M.  The  discussion  will  be  given  only  for  the  most  important 
practical  cases,  namely,  those  where  the  sections  are  rectangular. 

/  1     72 

For  these  —  equals  —?-  ,  and  formula  (4)  becomes, 

' 


In  this  bd*  must  vary  with  M  for  forms  of  uniform  strength 


ART.  38.   CANTILEVER  BEAMS   OF   UNIFORM   STRENGTH.         8 1 


Elevation 


For  a  cantilever  beam  with  a  load  P  at  the  end,  M  =  Px 
and  the  equation  becomes  \Sbd*  =  Px,  in  which  P  and  5  are 
constant.  If  the  breadth  be  taken  as  con- 
stant, d*  varies  with  x  and  the  profile  is 
that  of  a  parabola  whose  vertex  is  at  the 
load,  as  shown  in  Fig.  29.  The  equation  of 

the  parabola  is  d*  =  — —  x  from  which  d  may 

be  found  for  given  values  of  x.     The  walk-  Fig>  29> 

ing  beam  of  an  engine  is  often  made  approximately  of  this 
shape.  If  the  depth  of  the  cantilever  beam 
be  constant  then  b  varies  directly  as  x  and 
hence  the  plan  should  be  a  triangle  as  shown 
in  Fig.  30.  The  value  of  b  for  given  values 
of  P,  5,  d,  and  x  may  be  found  from  the 
6Px 


expression  b  = 


Sd'' 


Fig.  30. 


For  a  cantilever  beam  uniformly  loaded  with  w  per  linear 
unit  M  =  %wx*,  and  the  equation  becomes  \Sbd*  =  \wx*, 
in  which  w  and  .S  are  known.  If  the 
breadth  be  taken  as  constant  then  d  varies 
as  x  and  the  elevation  is  a  triangle,  as 
in  Fig.  31,  whose  depth  at  any  point  is 


If  however  the  depth  be 


Fig. 


ifi 

taken    constant,   then   b  =  -|—  x*  which  is  the  equation  of  a 

parabola  whose  vertex  is  at  the  free  end  of  the  cantilever  and 
whose  axis  is  perpendicular  to  it.  .Or  the  equation  may  be 
satisfied  by  two  parabolas  drawn  upon  opposite  sides  of  the 
center  line  as  shown  in  Fig.  32. 

The  vertical  shear  modifies  in  practice  the  shape  of  these 
forms  near  their  ends.     For  instance,  a  cantilever  beam  loaded 


82 


CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 


at  the  end  with  P  requires  a  cross-section  at  the  end  equal  to 

p 

-^r  where   Sc  is   the   working  shearing   strength.      This  cross- 

^c 

section  must  be  preserved  until  a  value  of 
x  is  reached,  where  the  same  value  of  the 
cross-section  is  found  from  the  moment. 


The  deflection  of  a  cantilever  beam  of 
uniform  strength  is  evidently  greater  than 
that  of  one  of  constant  cross-section,  since 
the  unit-stress  5  is  greater  throughout.  In 
any  case  it  may  be  determined  from  the  general  formula 
(5)  by  substituting  for  -M  and  /  their  values  in  terms  of  x,  in- 
tegrating twice,  determining  the  constants,  and  then  making 
x  equal  to  /  for  the  maximum  value  of  y. 

For  a  cantilever  beam  loaded  at  the  end  and  of  constant 
breadth,  as  in  Fig.  29,  formula  (5)  becomes, 

d*y       \2Px    _2 
dx*~  Ebd*  ~  E 


Integrating  this  twice  and   determining  the  constants,  as  in 
Art.  34,  the  equation  of  the  elastic  curve  is  found  to  be, 


In  this  make  x  =  /,  and  substitute  for  5  its  value  -—- ,  where 

bd* 

d^  is  the  depth  at  the  wall.     Then, 

spr 

~^bd~^ 
which  is  double  that  of  a  cantilever  beam  of  uniform  depth  d. 

For  a  cantilever  beam  loaded  at  the  end   and  of  constant 
depth,  formula  (5)  becomes, 

d*y  _  \2Px  _  2S 
~-~Ebd*~~~~Ed* 


ART.  39.      SIMPLE   BEAMS   OF   UNIFORM    STRENGTH.  83 

By  integrating  this  twice  and  determining  the  constants  as 
before,  the  equation  of  the  elastic  curve  is  found,  from  which 
the  deflection  is, 


which  is  fifty  per  cent  greater  than  for  one  of  uniform  section. 

Prob.  66.  A  cast  iron  cantilever  beam  of  uniform  strength  is 
to  be  4  feet  long,  3  inches  in  breadth  and  to  carry  a  load  of 
15  ooo  pounds  at  the  end.  Find  the  proper  depths  for  every 
foot  in  length,  using  3  ooo  pounds  per  square  inch  for  the  hori- 
zontal unit-stress,  and  4000  pounds  per  square  inch  for  the 
shearing  unit-stress. 


ART.  39.    SIMPLE  BEAMS  OF  UNIFORM  STRENGTH. 

In  the  same  manner  it  is  easy  to  deduce  the  forms  of  uni- 
form strength  for  simple  beams  of  rectangular  cross-section. 

For  a  load  at  the  middle  and  breadth  constant,  M  =  \Px> 

<>p 
and  hence,  ±Sbd*  =  \Px.     Hence  d*  —  ^—  x,  from  which  values 

00 

of  d  may  be  found  for  assumed  values  of  x.  Here  the  profile 
o.f  the  beam  will  be  parabolic,  the  vertex  being  at  the  support, 
and  the  maximum  depth  under  the  load. 

For  a  load  at  the  middle  and  depth  constant,  M  —  \Px  as 

\P 

before.     Hence  b  =  -FT^  and  the  plan  must  be  triangular  or 

Ow 

lozenge  shaped,  the  width  uniformly  increasing  from  the  sup- 
port to  the  load. 


For  a  uniform  load  and  constant  breadth,  M  =  %wlx  — 
and  hence,  ^/8=  ^-  (Ix  —  x*\  and  the  profile  of  the  beam  must 
be  elliptical,  of  preferably  a  half-ellipse. 


84  CANTILEVER  BEAMS  AND  SIMPLE  BEAMS.      CH.  III. 

*J*70) 

For  a  uniform  load  and  constant  depth,  b  =  -^—  (Ix  —  x*)  and 

ow 

hence  the  plan  should  be  formed  of  two  parabolas  having  their 
vertices  at  the  middle  of  the  span. 

The  figures  for  these  four  cases  are  purposely  omitted,  in 
order  that  the  student  may  draw  them  for  himself  ;  if  any  dif- 
ficulty be  found  in  doing  this  let  numerical  values  be  assigned 
to  the  constant  quantities  in  each  equation,  and  the  variable 
breadth  or  depth  be  computed  for  different  values  of  x. 

In  the  same  manner  as  in  the  last  article,  it  can  be  shown 
that  the  deflection  of  a  simple  beam  of  uniform  strength  loaded 
at  the  middle  is  double  that  of  one  of  constant  cross-section  if 
the  breadth  is  constant,  and  is  one  and  one-half  times  as  much 
if  the  depth  is  constant. 

Prob.  67.  Draw  the  profile  for  a  cast  iron  beam  of  uniform 
strength,  the  span  being  8  feet,  breadth  3  inches,  load  at  the 
middle  30,000  pounds,  using  the  same  working  unit-stresses  as 
in  Prob.  66. 

Prob.  68.  Find  the  deflection  of  a  steel  spring  of  constant;, 
depth  and  uniform  strength  which  is  6  inches  wide  at  the  mid- 
dle, 52  inches  long,  and  loaded  at  the  middle  with  600  pounds, 
the  depth  being  such  that  the  maximum  fiber  stress  is  20  OOQ 
pounds  per  square  inch. 


ART.  40.         BEAMS   OVERHANGING  ONE  SUPPORT. 


__________  ?  ________  x--m— 


CHAPTER   IV. 
RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS. 

ART.  40.    BEAMS  OVERHANGING  ONE  SUPPORT. 

A  cantilever  beam  has  its  upper  fibers  in  tension  and  the 
lower  in  compression,  while  a  simple  beam  has  its  upper  fibers 
in  compression  and  the  lower  in  tension.  Evidently  a  beam 
overhanging  one  support, 
as  in  Fig.  33,  has  its  over-  / 
hanging  part  in  the  con-  & 
dition  of  a  cantilever, 
and  the  part  near  the 
other  end  in  the  condi- 
tion of  a  simple  beam. 
Hence  there  must  be  a 
point  i  where  the  stresses 
change  from  tension  to 
compression,  and  where 
the  curvature  changes  F»g.  33- 

from  positive  to  negative.  This  point  i  is  called  the  inflection 
point  ;  it  is  the  point  where  the  bending  moment  is  zero.  An 
overhanging  beam  is  said  to  be  subject  to  a  restraint  at  the 
support  beyond  which  the  beam  projects,  or,  in  other  words, 
there  is  a  stress  in  the  horizontal  fibers  over  that  support. 

Since  the  beam  has  but  two  supports,  its  reactions  may  be 
found  by  using  the  principle  of  moments  as  in  Art.  16.  Thus, 
if  the  distance  between  the  supports  be  /,  the  length  of  the 


86  RESTRAINED   AND  CONTINUOUS   BEAMS.  CH.  IV. 

overhanging  part  be  my  and  the  uniform  load  per  linear  foot  be 
w,  the  two  reactions  are, 

wl       wm*  n       wl  .   u'm* 

*>  =  T-    ^'  R,  =  -  +  wm^--, 

whose  sum  is  equal  to  the  total  load  wl  -\-  wm.  Here,  as  in 
all  cases  of  uniform  load,  the  lever  arms  are  taken  to  the  cen- 
ters of  gravity  of  the  portions  considered. 

When  the  reactions  have  been  found,  the  vertical  shear  at 
any  section  can  be  computed  by  Art.  17,  and  the  bending  mo- 
ment by  Art.  1  8,  bearing  in  mind  that  for  a  section  beyond  the 
right  support  the  reaction  R^  must  be  considered  as  a  force 
acting  upward.  Thus,  for  any  section  distant  x  from  the  left 
support, 

When  x  is  less  than  /,  When  x  is  greater  than  /, 

V=  R,  -  wx,  JS=R1  +  R,-  wx, 


The  curves  corresponding  to  these  equations  are  shown  on 
Fig-  33-  The  shear  curve  consists  of  two  straight  lines  ;  F—  R, 

r> 

when  x  =  o,  and  V  =  o  when  x  =  —  -  ;    at  the  right  support 

V  =Rl—wl  from  the  first  equation,  and  V=  Rl  +  Rt  —  wl 
from  the  second  ;  V  =  o  when  x  =  l-\-  m.  The  moment  curve 
consists  of  two  parts  of  parabolas  ;  M  =  o  when  x  =.  o,  M  is  a 

r> 

maximum  when  x  =  —  l-,  M  =  o  at  the  inflection  point  where 

w 

x  =  -  l,  M  has  its  negative  maximum  when  x  =  /,  and  M  =  o 

w 

when  x  =  /+  m.  The  diagrams  show  clearly  the  distribution 
of  shears  and  moments  throughout  the  beam. 

For  example,  if  I  =  20  feet,  m  =  10  feet,  and  w  =  40  pounds 
per  linear  foot,  the  reactions  are  Rt  =  300  and  R,  =  900  pounds. 


ART.  40.         BEAMS   OVERHANGING   ONE   SUPPORT.  8/ 

Then  the  point  of  zero  shear  or  maximum  moment  is  at 
x  —  7.5  feet,  the  inflection  point  at  x  =  15  feet,  the  maximum 
shears  are  -(-  300,  —  500,  and  -f-  400  pounds,  and  the  maximum 
bending  moments  are  -\-  II25  and  —  2  ooo  pound-feet.  Here 
the  negative  bending  moment  at  the  right  support  is  numeri- 
cally greater  than  the  maximum  positive  moment.  The  rela- 
tive values  of  the  two  maximum  moments  depend  on  the  ratio 
of  m  to  /;  if  m  =  o  there  is  no  overhanging  part  and  the  beam 
is  a  simple  one  ;  if  m  =  \l  the  case  is  that  just  discussed  ;  if 
m  =  I  the  reaction  R^  is  zero,  and  each  part  is  a  cantilever  beam. 

After  having  thus  found  the  maximum  values  of  Fand  M 
the  beam  may  be  investigated  by  the  application  of  formulas 
(3)  and  (4)  in  the  same  manner  as  a  cantilever  or  simple  beam. 
By  the  use  of  formula  (5)  the  equation  of  the  elastic  curve  be- 
tween the  two  supports  is  found  to  be, 


From  this  the  maximum  deflection  for  any  particular  case  may 
be  determined  by  putting  -~-  equal  to  zero,  solving  for  x,  and 
then  finding  the  corresponding  value  of  y. 

If  concentrated  loads  be  placed  at  given  positions  on  the 
beam  the  reactions  are  found  by  the  principle  of  moments,  and 
then  the  entire  investigation  can  be  made  by  the  methods  above 
described. 

Prob.  69.  Three  men  carry  a  stick  of  timber,  one  taking  hold 
at  one  end  and  the  other  two  at  a  common  point.  Where 
should  this  point  be  so  that  each  may  bear  one  third  the 
weight  ?  Draw  the  diagrams  of  shears  and  moments. 

Prob.  70.  A  beam  20  feet  long  has  one  support  at  the  right 
end  and  one  support  at  5  feet  from  the  left  end.  At  the  left 
end  is  a  load  of  180  pounds,  and  at  6  feet  from  the  right  end 
is  a  load  of  125  pounds.  Find  the  reactions,  the  inflection 
point,  and  draw  the  shear  and  moment  diagrams. 


88  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

ART.  41.    BEAMS  FIXED  AT  ONE  END  'AND  SUPPORTED  AT 

THE  OTHER. 

A  beam  is  said  to  be  fixed  at  the  end  when  it  is  so  restrained 
in  a  wall  that  the  tangent  to  the  elastic  curve  at  the  wall  is 
horizontal.  Thus,  in  Fig.  33,  if  the  part  m  is  of  such  a  length 

that  the  tangent  over  the  right   sup- 

--*--  >!  i         i  i     port  is  horizontal,  the  part  /  is  in  the 

same  condition  as  a  beam  fixed  at 
one  end  and  supported  at  the  other. 
Fig.  34  shows  the  practical  arrange- 
ment of  such  a  beam,  the  left  sup- 
Flg*  34'  port  being  upon  the  same  level  as 

the  lower  side  of  the  beam  at  the  wall.  The  reactions  of  such 
a  beam  cannot  be  determined  by  the  principles  of  statics  alone, 
but  the  assistance  of  the  equation  of  the  elastic  curve  must  be 
invoked. 

Case  I.  For  a  uniform  load  over  the  whole  beam,  as  in  Fig. 
34,  let  R  be  the  reaction  at  the  left  end.  Then  for  any  section 
the  bending  moment  is  Rx  —  \wx*.  Hence  the  differential 
equation  of  the  elastic  curve  is, 


Integrate  this  once  and  determine  the  constant  from  the  neces- 

dv 
sary  condition  that  ~  =  o  when  x  =  I.     Integrate  again  and 

find  the  constant  from  the  fact  that  y  =  o  when  x  =  o.     Then, 


Here  also  y  =  o  when  x  =  /,  and  therefore  R  = 

The  moment  at  any  point  now  is  M  =  \wlx  —  %wx*,  and  by 
placing  this  equal  to  zero  it  is  seen  that  the  point  of  inflection 
is  at  x  =  |/.  By  the  method  of  Art.  24  it  is  found  that  the 


ART.  41.  BEAMS   FIXED   AT   ONE   END.  89 


maximum  moments  are+yl-g-w/8  and  —  \wl*,  and  that  the 
distribution  of  moments  is  as  represented  in  Fig.  34. 

dv 
The  point  of  maximum  deflection  is   found  by  placing  -*- 

equal  to  zero  and  solving  for  x.  Thus  8*8  —  glx*  +  /3  =  o, 
one  root  of  which  is  x  =  +0.42157,  and  this  inserted  in  the 
value  of  y  gives, 

J  =  0.0054  j£, 

for  the  value  of  the  maximum  deflection. 

Case  II.  For  a  load  at  the  middle  it  is  first  necessary  to  con- 
sider that  there  are  two  elastic  curves  having  a  common  ordi- 
nate  and  a  common  tangent  under  the  load,  since  the  expres- 
sions for  the  moment  are  different  on  opposite  sides  of  the 
load.  Thus,  taking  the  origin  as  usual  at  the  supported  end, 

On  the  left  of  the  load, 

(.) 

(*)  £/£ 


On  the  right  of  the  load  the  similar  equations  are, 
(a)'  EI^^Rx-P^-V), 

(6)'  E/& 

(c)' 


To  determine  the  constants  consider  in  (*:)  that  y  =  o  when 
x  =  o  and  hence  that  C,  =  o.  In  (&)'  the  tangent  —  =  o  when 
x  =  /  and  hence  C^=  —  \Rr.  Since  the  curves  have  a  common 


90  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

tangent  under  the  load  (b)  =  (&)'  for  x  =%!,  and  thus  the  value 
of  Cv  is  found.  Since  the  curves  have  a  common  ordinate 
under  the  load  (c)  =  (c)'  when  x  =  £/,  and  thus  Ct  is  found. 

Then, 

_  RJ_       Prx       Rl*x 
(c)  Ely-      g-+     g  2     , 

Rx*      Px*       Plx*      Rl*x    ,  Pr 
(c)'  Ely--    -  +  ~+' 


From  the  second  of  these  the  value  of  the  reaction  is  R  = 


The  moment  on  the  left  of  the  load  is  now  M  =  -£^Px,  and 
that  on  the  right  M  =  —  \%Px  +  \PL  The  maximum  posi- 
tive moment  obtains  at  the  load  and  its  value  is  ^Pl-  The 
maximum  negative  moment  occurs  at  the  wall,  and  its  value  .is 
-f^Pl.  The  inflection  point  is  at  x  =  -f^l.  The  deflection  un- 
der the  load  is  readily  found  from  (c)  by  making  x  —  £/.  The 
maximum  deflection  occurs  at  a  less  value  of  x,  which  may  be 
found  by  equating  the  first  derivative  to  zero. 

Case  III.  For  a  load  at  any  point  whose  distance  from  the 
left  support  is  kl,  the  following  results  may  be  deduced  by  a 
method  exactly  similar  to  that  of  the  last  case. 

Reaction  at  supported  end  =  %P(2  —  $k  -f-  k?). 

Reaction  at  fixed  end  =  \P(^k  —  k*\ 

Maximum  positive  moment  =  \Plk(2  —  ^k-^-k*}. 

Maximum  negative  moment  =  %Pl(k  —  £3). 

The  absolute  maximum  deflection  occurs  under  the  load  when 


Prob.  71.  Draw  the  diagrams  of  shears  and  moments  for 
a  load  at  the  middle,  taking  P  •=.  600  pounds  and  /=  12  feet. 

Prob.  72.  Find  the  position  of  load  Pwhich  gives  the  maxi- 
mum positive  moment.  Find  also  the  position  which  gives  the 
maximum  negative  moment. 


ART.  42.    BEAMS  OVERHANGING  BOTH  SUPPORTS.  QI 

ART.  42.    BEAMS  OVERHANGING  BOTH  SUPPORTS. 

When  a  beam  overhangs  both  supports  the  bending  moments 
for  sections  beyond  the  supports  are  negative,  and  in  general 
between  the  supports  there  will  be  two  inflection  points.  If 
the  lengths  m  and  n  be 
equal  the  reactions  will  be  ^m-^~ 


equal   under    uniform    load, 

each  being  one  half  of  the 

total     load.      In    any  case, 

whatever  be  the    nature  of 

the  load,  the  reactions    may 

be  found  by  the  principle  of  moments  (Art.  16),  and  then  the 

vertical  shears  and  bending  moments  may  be  deduced  for  all 

sections,  after  which  the  formulas  (3)  and  (4)  can  be  used  for 

any  special  problem. 

Under  a  uniformly  distributed  load,  and  m  =  n,  which  is  the 
most  important  practical  case,  each  reaction  is  wm  -\-  %wl,  the 
maximum  shears  at  the  supports  are  wm  and  \wl,  the  maxi- 
mum moment  at  the  middle  is  +  w(^r  —  %m*),  the  maximum 
moment  at  each  support  is  —  $wm*,  and  the  inflection  points 


are  distant  J  V  T  —  4m3  from  the  middle  of  the  beam.  Fig.  35 
shows  the  distribution  of  moments  for  this  case.  If  m  =  o,  the 
beam  is  a  simple  one  ;  if  /  =  o,  it  consists  of  two  cantilever 
beams. 

Prob.  73.  If  m  =  n  in  Fig.  35,  find  the  ratio  of  /  to  m  in 
order  that  the  maximum  positive  moment  may  numerically 
equal  the  maximum  negative  moment. 

Prob.  74.  A  beam  30  feet  long  has  one  support  at  5  feet 
from  the  left  end,  and  the  other  support  at  10  feet  from  the 
right  end.  At  each  end  there  is  a  load  of  156  pounds  and 
half-way  between  the  supports  there  is  a  load  of  344  pounds. 
Construct  shear  and  moment  diagrams. 


92  RESTRAINED  AND  CONTINUOUS  BEAMS.         CH.  IV. 

ART.  43.    BEAMS  FIXED  AT  BOTH  ENDS. 

If,  in  Fig.  35,  the  distances  m  and  n  be  such  that  the  elastic 
curve  over  the  supports  is  horizontal  the  central  span  /  is 
said  to  be  a  beam  fixed  at  both  ends.  The  lengths  m  and  n 
which  will  cause  the  curve  to  be  horizontal  at  the  support  can 
be  determined  by  the  help  of  the  elastic  curve.  For  uniform 
load  n  =  m  and  the  bending  moment  at  any  section  in  the 
span  /  distant  x  from  the  left  support  is, 


M  =  (wm  +  %wl)x  —  %w(m  -f- 
which  reduces  to  the  simpler  form, 
M  =  Mf 


in  which  M'  represents  the  unknown  bending  moment  —  \wn? 
at  the  left  support. 

Again,  for  a  single  load  P  at  the  middle  of  /in  Fig.  35  the 
elastic  curve  can  be  regarded  as  kept  horizontal  at  the  left  sup- 
port by  a  load  Q  at  the  end  of  the  distance  m.  Then  the  bend- 
ing moment  at  any  section  distant  x  from  the  left  support,  and 
between  that  support  and  the  middle,  is, 


which  reduces  to, 


in  which  M  '  denotes  the  unknown  moment  —  Qm  at  the  left 
support.  The  problem  of  finding  the  bending  moment  at  any 
section  hence  reduces  to  that  of  determining  M'  the  moment 
at  the  support. 

Case  I.  For  a  uniform  load  the  general  equation  of  the  elastic 
curve  now  is, 


ART.  43.  BEAMS    FIXED   AT   BOTH    ENDS.  93 

Integrating  this  twice,  making  -jj-  =  o  when  x  =  o  and  also 


when  x  =  /,  the  value  of  M' 


\ 

^  ---  ^i 


is    found    to    be  -       — ,   and 


the  equation   of  the  elastic  ^tll^ 

curve  becomes,  m? 

2±EIy  =  w(-l*x*  +  2lx*  -  S).  Fi*-  36- 

From  this  the  maximum  deflection  is  found  to  be, 

~ 


The  inflection  points  are  located  by  making  M  =  o,  which 
gives  x  —  4/±  /A  /  —  •  The  maximum  positive  moment  is  at 

wr 

the  middle  and  its  value  is   -   -  ;    accordingly  the  horizontal 

24 

stress  upon  the  fibers  at  the  middle  of  the  beam  is  one  half 
that  at  the  ends.  The  vertical  shear  at  the  left  end  is  %w/,  at 
the  middle  o,  and  at  the  right  end  — 


Case  II.  For  a  load  at  the  middle  the  general  equation  of 
the  elastic  curve  between  the  left  end  and  the  load  is, 


and  in  a  similar  manner  to  that 

of  the  last  case  it  is  easy  to  find 

that  the  maximum  negative  mo-  Fi&-  37. 

ments  are  -J/V,  that  the  maximum  positive  moment  is  -J/Y,  that 

the  inflection  points  are  half-way  between  the  supports  and  the 

pr 

load,  and  that  the  maximum  deflection  is  -  -=-=:  - 

192^7 

:     Case  III.  For  load  P  at  a  distance  kl  from  the  left  end  let 


94  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

M'  and  V  denote  the  unknown  bending  moment  and  vertical 
shear  at  that  end.  Then  on  the  left  of  the  load, 

M=M'+  V'x, 
and  on  the  right  of  the  load 

M=  M'  +  V'x  —  P(x  -  kl). 

By  inserting  these  in  the  general  formula  (5),  integrating  each 
twice  and  establishing  sufficient  conditions  to  determine  the 
unknown  M'  and  V  and  also  the  constants  of  integration,  the 
following  results  may  be  deduced, 

Shear  at  left  end  =  P(\  —  $tf  +  2>P), 

Shear  at  right  end  =  Pk\^  —  2k). 

Moment  at  left  end  =  —  Plk(\  —  2k  +  JP), 

Moment  at  right  end  =  —  Plk\\  —  k\ 

Moment  under  load  =  +  Plk\2  —  4&  -\-  2&). 

If  k  •=.  \  the  load  is  at  the  middle  and  these  results  reduce  to 
the  values  found  in  Case  II. 

Prob.  75.  Show  from  the  .results  above  given  for  Case  III 

kl 

that   the   inflection   points  are   at  the   distances  -  -   and 

i  +  2k 

th  left  end 


Prob.  76.  What  medium-steel  I  beam  is  required  for  a  span 
of  24  feet  to  support  a  uniform  load  of  25  ooo  pounds,  the  ends 
being  merely  supported?  What  one  is  needed  when  the  ends 
are  fixed? 

ART.  44.    COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS. 

As  the  maximum  moments  for  restrained  beams  are  less 
than  for  simple  beams  their  strength  is  relatively  greater.  This 
was  to  be  expected,  since  the  restraint  produces  a  negative 
bending  moment  and  lessens  the  deflection  which  would  other- 


ART.  44.  COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS.  95 


wise  occur.  The  comparative  strength  and  stiffness  of  canti- 
levers and  simple  beams  is  given  in  Art.  37.  To  these  may 
now  be  added  four  cases  from  Arts.  41  and  43,  and  the  follow- 
ing table  be  formed,  in  which  W  represents  the  total  load, 
whether  uniform  or  concentrated. 


Beams  of  Uniform  Cross-section. 

Maximum 
Moment. 

Maximum 
Deflection. 

Relative 
Strength. 

Relative 
Stiffness. 

I    Wl* 

Cantilever,  load  at  end 

Wl 

I 

I 

3  ~£I 

I    Wl* 

Cantilever,  uniform  load 

l  yyi 

2 

2| 

8    £~f 

8 

Simple  beam,  load  at  middle 

i  f^/y 

48  ~EI 

4 

16 

Simple  beam  uniformly  loaded 

\Wl 

384   El 

8 

25f 

Beam  fixed  at  one  end,  supported  at 

Wl* 

other,  load  near  middle 

0.192^7 

o.o,8a_ 

5-2 

18.3 

Beam  fixed  at  one  end,  supported  at 
other,  uniform  load 

\Wl 

W7» 
0.0054-^ 

8 

62 

Beam   fixed   at    both    ends,    load   at 

i     Wl* 

middle 

i  Wl 

8 

64 

192    El 

Beam   fixed   at    both   ends,   uniform 

i     Wl* 

load 

T^f  W/ 

12 

128 

p4^7 

This  table  shows  that  a  beam  fixed  at  both  ends  and  uni- 
formly loaded  is  one  and  one-half  times  as  strong  and  five  times 
as  stiff  as  a  simple  beam  under  the  same  load.  The  advantage 
of  fixing  the  ends  is  hence  very  great. 

Prob.  77.  Prove,  for  a  uniformly  loaded  beam  with  equal 
overhanging  ends,  that  the  deflection  at  the  middle  is  given  by 

the  formula  '  ($/'  —  24^'). 

354/i/ 

'Prob.  78.  Find  the  deflection  of  a  g-inch  I  beam  of  6  feet 
span  and  fixed  ends  when  loaded  at  the  middle  so  that  the 
tensile  and  compressive  stresses  at  the  dangerous  section  are 
1 6  800  pounds  per  square  inch. 


96  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

ART.  45.    GENERAL  PRINCIPLES  OF  CONTINUITY. 

A  continuous  beam  is  one  supported  upon  several  points  in 
the  same  horizontal  plane.  A  simple  beam  may  be  regarded 
as  a  particular  case  of  a  continuous  beam  where  the  number  of 
supports  is  two.  The  ends  of  a  continuous  beam  are  said  to 
be  free  when  they  overhang,  supported  when  they  merely  rest 
on  abutments,  and  restrained  when  they  are  horizontally  fixed 
in  walls. 

The  general  principles  of  the  preceding  chapter  hold  good 
for  all  kinds  of  beams.  If  a  plane  be  imagined  to  cut  any 
beam  at  any  point  the  laws  of  Arts.  19  and  20  apply  to  the 
stresses  in  that  section.  The  resisting  shear  and  the  resisting 
moment  for  that  section  have  the  values  deduced  in  Art.  21 
and  the  two  fundamental  formulas  for  investigation  are, 

(3)  S,A  =  V, 

(4)  ^=M 

Here  5S  is  the  vertical  shearing  unit-stress  in  the  section,  and 
»S  is  the  horizontal  tensile  or  compressive  unit-stress  on  the 
fiber  most  remote  from  the  neutral  axis ;  c  is  the  shortest  dis- 
tance from  that  fiber  to  that  axis ;  /  the  moment  of  inertia, 
and  A  the  area  of  the  cross-section.  V  is  the  vertical  shear  of 
the  external  forces  on  the  left  of  the  section,  and  M  is  the 
bending  moment  of  those  forces  with  reference  to  a  point  in 
the  section.  For  any  given  beam  evidently  5S  and  S  may  be 
found  for  any  section  as  soon  as  V  and  M  are  known. 

The  general  equation  of  the  elastic  line,  deduced  in  Art.  33, 
is  also  valid  for  all  kinds  of  beams.  It  is, 

(5) 
UJ 


ART.  45.       GENERAL   PRINCIPLES   OF   CONTINUITY.  97 

where  x  is  the  abscissa  and  y  the  ordinate  of  any  point  of  the 
elastic  curve,  M  being  the  bending  moment  for  that  section, 
and  E  the  coefficient  of  elasticity  of  the  material. 

The  vertical  shear  V  is  the  algebraic  sum  of  the  external 
forces  on  the  left  of  the  section,  or,  as  in  Art.  17, 

V  •=  Reactions  on  left  of  section  minus  loads  on  left  of  section. 

For  simple  beams  and  cantilevers  the  determination  of  V 
for  any  special  case  was  easy,  as  the  left  reaction  could  be 
readily  found  for  any  given  loads.  For  continuous  beams, 
however,  it  is  not,  in  general,  easy  to  find  the  reactions,  and 
hence  a  different  method  of  determining  Fis  necessary.  Let 
Fig.  38  represent  one  span  of  a  continuous  beam.  Let  V  be 
the  vertical  shear  for  any  L  j 

*  h^-  —  .*»•»  -X-  __  ^_  ^4 

section  at  the  distance  x  __  _Jpi 


from  the  left  support,  and    )  \ 

V  the  vertical  shear  at  a          ^r' 

section  infinitely  near  to  Fig.  38. 

the  left  support.     Also  let  2Pt  denote  the  sum  of  all  the  con- 

centrated loads  on  the  distance  x,  and  wx  the  uniform  load. 

Then  because  V  is  the  algebraic  sum  of  all  the  vertical  forces 

on  its  left,  the  definition  of  vertical  shear  gives, 

(6)  V=  V  -wx-^P^ 

Hence  V  can  be  determined  as  soon  as  V  is  known. 

The  bending  moment  M  is  the  algebraic  sum  of  the  mo- 
ments of  the  external  forces  on  the  left  of  the  section  with  ref- 
erence to  a  point  in  that  section,  or,  as  in  Art.  18, 

M  -=.  moments  of  reactions  minus  moments  of  loads. 

For  the  reason  just  mentioned  it  is  in  general  necessary  to  de- 
termine M  for  continuous  and  restrained  beams  by  a  different 
method.  Let  M'  denote  the  bending  moment  at  the  left  sup- 
port of  any  span  as  in  Fig.  38,  and  M"  that  at  the  right  sup- 


98  RESTRAINED  AND   CONTINUOUS   BEAMS.  CH.  IV. 

port,  while  M  is  the  bending  moment  for  any  section  distant  x 
from  the  left  support.  Let  Pl  be  any  concentrated  load  upon 
the  space  x  at  a  distance  £/  from  the  left  support,  k  being  a 
fraction  less  than  unity,  and  let  w  be  the  uniform  load  per  lin- 
ear unit.  Let  V  be  the  resultant  of  all  the  vertical  forces  on 
the  left  of  a  section  in  the  given  span  infinitely  near  to  the 
left  support,  and  let  m  be  the  distance  of  the  point  of  applica- 
tion of  that  resultant  from  that  support.  Then  the  definition 
of  bending  moment  gives, 


But  Vm  is  the    unknown   bending   moment  M'  at  the    left 
support.     Hence 

(7)        M=Mf+V'x-%wx*  —  2*Pl(x-kl), 

from  which  M  may  be  found  for  any  section  as  soon  as  M'  and 
V  have  been  determined. 

The  vertical  shear  V  at  the  support  may  be  easily  found  if 
the  bending  moments  M'  and  M"  be  known.  Thus  in  equa- 
tion (7)  make  x  =  /,  then  M  becomes  M",  and  hence, 


The  whole  problem  of  the  discussion  of  restrained  and  con- 
tinuous beams  hence  consists  in  the  determination  of  the  bend- 
ing moments  at  the  supports.  When  these  are  known  the 
values  of  M  and  V  may  be  determined  for  every  section,  and 
the  general  formulas  (3),  (4),  and  (5)  be  applied  as  in  Chapter 
III,  to  the  investigation  of  questions  of  strength  and  deflec- 
tion. The  formulas  (6),  (7),  and  (8)  apply  to  cantilever  and 
simple  beams  also.  For  a  simple  beam  M'  —  M"  =  o,  and 
V  =  R.  For  a  cantilever  beam  M'  =  o  for  the  free  end,  and 
M"  is  the  moment  at  the  wall. 

The  relation  between  the  bending  moment  and  the  vertical 


ART.  46.    PROPERTIES  OF  CONTINUOUS  BEAMS.         99 

shear  at  any  section  is  interesting  and  important.  At  the  sec- 
tion x  the  moment  is  M  and  the  shear  is  V.  At  the  next  con- 
secutive section  x  -)-  dx  the  moment  is  M  -\-  dM,  which  may 
also  be  expressed  by  M-\-  Vdx.  Hence, 

dM 

~  dx' 

This  may  be  proved  otherwise  by  differentiating  (7)  and  com- 
paring with  (6).  From  this  it  is  seen  that  the  maximum  mo- 
ments occur  at  the  sections  where  the  shear  passes  through 
zero. 

Prob.  79.  A  bar  of  length  2/  and  weighing  w  per  linear 
unit  is  supported  at  the  middle.  Apply  formulas  (6)  and  (7) 
to  the  statement  of  general  expressions  for  the  moment  and 
shear  at  any  section  on  the  left  of  the  support,  and  also  at  any 
section  on  the  right  of  the  support. 

ART.  46.    PROPERTIES  OF  CONTINUOUS  BEAMS. 

The  theory  of  continuous  beams  presented  in  the  following 
pages  includes  only  those  with  constant  cross-section  having 
the  supports  on  the  same  level,  as  only  such  are  used  in  engin- 
eering constructions.  Unless  otherwise  stated,  the  ends  will 
be  supposed  to  simply  rest  upon  their  supports,  so  that  there 
can  be  no  moments  at  those  points.  Then  the  end  spans  are 
somewhat  in  the  _  h  _  la  _  h  _  i* 
condition  of  A?  A3  £4  A« 


a  beam  with  one 


overhanging  end,     Ulflflllft^ 
and   the   other 

spans   somewhat  Fig.  39- 

in  the  condition  of  a  beam  with  two  overhanging  ends.  At 
each  intermediate  support  there  is  a  negative  moment,  and  the 
distribution  of  moments  throughout  the  beam  will  be  as  repre- 
sented in  Fig.  39. 


IOO  RESTRAINED  AND   CONTINUOUS   BEAMS.  CH.  IV. 

As  shown  in  Art.  45,  the  investigation  of  a  continuous  beam 
depends  upon  the  determination  of  the  bending  moments  at 
the  supports.  In  the  case  of  Fig.  39  these  moments  being 
those  at  the  supports  2,  3,  and  4,  may  be  designated  Mt  ,  Mz  , 
and  M4  .  Let  V^  V^,  V9,  and  V^  denote  the  vertical  shear  at  the 
right  of  each  support.  The  first  step  is  to  find  the  moments 
MI  ,  M3  ,  and  M4  .  Then  from  formula  (8)  the  values  of  Vl  ,  F2  , 
F3  ,  and  F<  are  found,  and  thus  by  formula  (7)  an  expression  for 
the  bending  moment  in  each  span  may  be  written,  from  which 
the  maximum  positive  moments  may  be  determined.  Lastly, 
by  formulas  (3)  and  (4)  the  strength  of  the  beam  may  be  in- 
vestigated, and  by  (5)  its  deflection  at  any  point  be  deduced. 

For  example,  let  the  beam  in  Fig.  39  be  regarded  as  of  four 
equal  spans  and  uniformly  loaded  with  w  pounds  per  linear 
unit.  By  a  method  to  be  explained  in  the  following  articles  it 
may  be  shown  that  the  bending  moments  at  the  supports  are, 


From  formula  (8)  the  vertical  shears  at  the  right  of  the  several 
supports  are, 

etc. 


And  from  (6)  those  on  the  left  of  the  supports  2,  3,  4,  etc.,  are 
found  to  be,  —  %%wl,  —  \\wl,  —  ^wl,  etc.  From  formula  (7) 
the  general  expressions  for  the  bending  moments  now  are, 

For  first  span,  M= 

For  second  span,  M=  — 

For  third  span,  M=— 

'   For  fourth  span,  M  =  — 

From  each  of  these  equations  the  inflection  points  may  be 
found  by  putting  M=o,  and  the  point  of  maximum  positive 

moment    by   putting     -—-  =  0.     The  maximum  positive    mo- 


ART.  46.         PROPERTIES   OF   CONTINUOUS   BEAMS.  IOI 

ments  are  found  to  have  the  following  values, 

TV2*W.         Ttfcr*"'1.         iHiw'1.         and        T^W/'. 

For  any  particular  case  the  beam  may  now  be  investigated  by 
formulas  (3)  and  (4). 

The  reactions  at  the  supports  are  not  usually  needed  in  the 
discussion  of  continuous  beams,  but  if  required  they  may 
easily  be  found  from  the  adjacent  shears.  Thus  for  the  above 
case, 


=  ftwl,  etc., 
and  the  sum  of  these  is  equal  to  the  total  load 

The  equation  of  the  elastic  curve  in  any  span  is  deduced  by 
inserting  in  (5)  for  M  its  value  and  integrating  twice.  When 

dv 
x  =  o,  the  tangent  -~  is  the  tangent  of  the  inclination  at  the 

left  support,  and  when  x  =  I  it  is  the  tangent  of  the  inclination 
at  the  right  support.  When  x  =  o,  and  also  when  x  =  /,  the 
ordinate  y  =  o,  and  from  these  conditions  the  two  unknown 
tangents  may  be  found.  In  general  the  maximum  deflection 
in  any  span  of  a  continuous  beam  will  be  found  intermediate 
in  value  between  those  of  a  simple  beam  and  a  restrained 
beam. 

In  the  following  pages  continuous  beams  will  only  be  inves- 
tigated for  the  case  of  uniform  load.  The  lengths  of  the  spans 
however  may  be  equal  or  unequal,  and  the  load  per  linear  foot 
may  vary  in  the  different  spans. 

Prob.  80.  In  a  continuous  beam  of  three  equal  spans  the 
negative  bending  moments  at  the  supports  are  -fawl*.  Find 
the  inflection  points,  the  maximum  positive  moments  and  the 
reactions  of  the  supports. 


IO2  RESTRAINED  AND  CONTINUOUS  BEAMS.      CH.  IV. 

ART.  47.    THE  THEOREM  OF  THREE  MOMENTS. 

Let  the  figure  represent  any  two  adjacent  spans  of  a  continu- 
ous beam  whose  lengths  are  I'  and  I"  and  whose  uniform  loads 
per  linear  foot  are  w'  and  w"  respectively.  Let  M'  ,  M",  and 

M'"     represent     the 
three     unknown    mo- 

af  tVi^  cnr»r»r»rfc 
clL  LJnC  SUL)L)Ol  Lz>» 

&*v>  ^jv"  Let  V  and  V"  be  the 

vertical  shears  at  the 

Fig-  4°"  right  of  the  first  and 

second  supports.  Then,  for  any  section  distant  x  from  the 
left  support  in  the  first  span,  the  moment  is, 

M=M'  +  V'x-\wx\ 

If  this  be  inserted  in  the  general  formula  (5)  and  integrated 
twice  and  the  constants  determined  by  the  condition  that 
y  =  o  when  x  —  o  and  also  when  x  —  /,  the  value  of  the  tan- 
gent of  the  angle  which  the  tangent  to  the  elastic  curve  at  any 
section  in  the  first  span  makes  with  the  horizontal  is  found 
to  be, 

dy  _  \2M\2x  - 
dx 

Similarly  if  the  origin  be  taken  at  the  next  support  the  value 
of  the  tangent  of  inclination  at  any  point  in  the  second  span  is> 

dy_  _ 


dx  24^7 

Evidently  the  two  curves  must  have  a  common  tangent  at  the 
support.  Hence  make  x  =  I'  in  the  first  of  these  and  x  =  o 
in  the  second  and  equate  the  results,  giving, 

"-*e//"=  -  \2M"l"  -    V"l'n       w"l"\ 


ART.  48.     CONTINUOUS  BEAMS  WITH  EQUAL  SPANS.         103 

Let  the  values  of   V  and  V"  be  expressed  by  (8)  in  terms  of 
Mf,  M" ,  and  M"' ,  and  the  equation  reduces  to, 

7f//"  7J/'I"* 

(g)    M'l'  +  2M"(r  +  t")  +  M'"i"  =  — -  -, 

4  4 

which  is  the  theorem  of  three  moments  for  continuous  beams 
uniformly  loaded. 

If  the  spans  are  all  equal  and  the  load  uniform  throughout, 
this  reduces  to  the  simpler  form, 

M'  +  ^M"  +  M"'  =  —  —. 

In  any  continuous  beam  of  s  spans  there  are  s  -f-  I  supports 
and  s  —  I  unknown  bending  moments  at  the  supports.  For 
each  of  these  supports  an  equation  of  the  form  of  (g)  may  be 
written  containing  three  unknown  moments.  Thus  there  will 
be  stated  s  —  I  equations  whose  solution  will  furnish  the  values 
of  the  s  —  I  unknown  quantities. 

Prob.  8 1.  A  simple  wooden  beam  one  inch  square  and  15 
inches  long  is  uniformly  loaded  with  100  pounds.  Find  the 
angle  of  inclination  of  the  elastic  curve  at  the  supports. 


ART.  48.    CONTINUOUS  BEAMS  WITH  EQUAL  SPANS. 

Consider  a  continuous  beam  of  five  equal  spans  uniformly 
loaded.  Let  the  supports  beginning  on  the  left  be  numbered 
I,  2,  3,  4,  5,  and  6.  From  the  theorem  of  three  moments  an 
equation  may  be  written  for  each  of  the  supports  2,  3,  4,  and 
5  ;  thus, 


104  RESTRAINED   AND   CONTINUOUS   BEAMS.  ClI.  IV. 

Since  the  ends  of  the  beam  rest  on  abutments  without  restraint 
Ml  =  M6  —  o.  Hence  the  four  equations  furnish  the  means 
of  finding  the  four  moments  M^,  M3,  Mt,  M6.  The  solution 
may  be  abridged  by  the  fact  that  M..  =  M6,  and  M3  =  M^ 
which  is  evident  from  the  symmetry  of  the  beam.  Hence, 


From  formula  (8)  the  shears  at  the  right  of  the  supports  are, 


etc. 

From  (7)  the  bending  moment  at  any  point  in  any  span  may 
now  be  found  as  in  Art.  46,  and  by  (3),  (4),  and  (5)  the  complete 
investigation  of  any  special  case  may  be  effected. 

In  this  way  the  bending  moments  at  the  supports  for  any 
number  of  equal  spans  can  be  deduced.  The  following  tri- 
angular table  shows  their  values  for  spans  as  high  as  seven  in 
number.  In  each  horizontal  line  the  supports  are  represented 
by  squares  in  which  are  placed  the  coefficients  of  —  wl*.  For 
example,  in  a  beam  of  3  spans  there  are  four  supports  and  the 
bending  moments  at  those  supports  are  o,  —  -fowl*,  — 
and  o. 


Equal  Spans. 

Moments. 


Fig.  41. 

The  vertical  shears  at  the  supports  are  also  shown  in  the 
following  table  for  any  number  of  spans  up  to  5.  The  space 
representing  a  support  shows  in  its  left-hand  division  the  shear 
on  the  left  of  that  support  and  in  its  right-hand  division 


ART.  48.    CONTINUOUS  BEAMS  WITH  EQUAL  SPANS. 


105 


the  shear  on  the  right.  The  sum  of  the  two  shears  for  any 
support  is,  of  course,  the  reaction  of  that  support.  For  ex- 
ample, in  a  beam  of  five  equal  spans  the  reaction  at  the  second 
support  is 


Fig.  42. 

It  will  be  seen  on  examination  that  the  numbers  in  any 
oblique  column  of  these  tables  follow  a  certain  law  of  increase 
by  which  it  is  possible  to  extend  them,  if  desired,  to  a  greater 
number  of  spans  than  are  here  given. 

As  an  example,  let  it  be  required  to  select  a  I  beam  to  span 
four  openings  of  8  feet  each,  the  load  per  span  being  14000 
pounds  and  the  greatest  horizontal  stress  in  any  fiber  to  be 
12000  pounds  per  square  inch.  The  required  beam  must 
satisfy  formula  (4),  or, 

7=      M 

C~~   12  000 

where  M  is  the  maximum  moment.  From  the  table  it  is  seen 
that  the  greatest  negative  moment  is  that  at  the  second  sup- 
port, or  ~w/*.  The  maximum  positive  moments  are, 

2o 


V* 

max  M  =  —  = 


For  fust  span, 

For  second  span,    max  M  = 


77» 

r,  +  —  = 

2W 


106  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

The  greatest  value  of  M  is  hence  at  the  second  support.    Then, 

/  =  3  X  14  OOP  X  8  X  12  =  I2 
c  28  X  12  ooo 

and  from  the  table  in  Art.  30  it  is  seen  that  a  light  7-inch  beam 
will  be  required. 

Prob.  82.  Draw  the  diagram  of  shears  and  the  diagram  of 
moments  for  the  case  of  three  equal  spans  uniformly  loaded. 

Prob.  83.  Find  what  I  beam  is  required  to  span  three  open- 
ings of  12  feet  each,  the  load  on  each  span  being  6  ooo  pounds, 
and  the  greatest  value  of  5  to  be  12000  pounds  per  square 
inch. 

ART.  49.    CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS. 

As  the  first  example,  consider  two  spans  whose  lengths  are 
/,,/,,  and  whose  loads  per  linear  unit  are  wl  and  w^  .  The 
theorem  of  three  moments  in  (9)  then  reduces  to, 


and  hence  the  bending  moment  at  the  middle  support  is, 

M 


From  this  the  reaction  at  the  left  support  may  be  found  by  (8) 
and  the  bending  moment  at  any  point  by  (7). 

Next  consider  three  spans  whose  lengths  are  /x  ,  /,  ,  and  /3  , 
loaded  uniformly  with  wl  ,  w^  ,  w3  .  The  bending  moments  at 
the  second  and  third  supports  are  M9  and  M3  .  Then  from  (9), 


and  the  solution  of  these  gives  the  values  of  M^  and  Ma  .  A 
very  common  case  is  that  for  which  ^  =  /,  /,  =  /8  =  «/,  and 
wl=w9  =  w,=  w.  For  this  case  the  solution  gives, 


ART.  50.      REMARKS   ON   THE   THEORY   OF  FLEXURE.  IO/ 

Here  if  n  =  I,  these  two  moments  become  —  -fowl*,  as  also 
shown  in  the  last  article. 

Whatever  be  the  lengths  of  the  spans  or  the  intensity  of  the 
loads,  the  theorem  of  three  moments  furnishes  the  means  of 
finding  the  bending  moments  at  the  supports.  Then  from  (8), 
(7),  and  (6)  the  vertical  shears  and  bending  moments  at  every 
section  may  be  computed.  Finally,  if  the  material  be  not 
strained  beyond  its  elastic  limit,  formula  (5)  may  be  used  to 
determine  the  deflection,  while  (4)  investigates  the  strength  of 
the  beam. 

Prob.  84.  A  continuous  beam  of  three  equal  spans  is  loaded 
only  in  the  middle  span.  Find  the  reactions  of  the  end  sup- 
ports due  to  this  load. 

Prob.  85.  A  heavy  1 2-inch  I  beam  of  36  feet  length  covers 
four  openings,  the  two  end  ones  being  each  8  feet  and  the 
others  each  10  feet  in  span.  Find  the  maximum  moment  in 
the  beam.  Then  determine  the  load  per  linear  foot  so  that  the 
greatest  horizontal  unit-stress  may  be  12000  pounds  per  square 
inch. 


ART.  50.    REMARKS  ON  THE  THEORY  OF  FLEXURE. 

The  theory  of  flexure  presented  in  this  and  the  preceding 
chapter  is  called  the  common  theory,  and  is  the  one  universally 
adopted  for  the  practical  investigation  of  beams.  It  should  not 
be  forgotten,  however,  that  the  axioms  and  laws  upon  which 
it  is  founded  are  only  approximate  and  not  of  an  exact  nature 
like  those  of  mathematics.  Laws  (A)  and  (B)  for  instance  are 
true  as  approximate  laws  of  experiment,  but  probably  not  as 
exact  laws  of  science.  Law  (G)  has  been  established  by  the 
observed  fact  that  a  vertical  line,  drawn  upon  the  side  of  the 
beam  before  flexure,  remains  a  straight  line  after  flexure,  even 
when  the  elastic  limit  of  the  material  is  exceeded. 


IO8  RESTRAINED   AND   CONTINUOUS   BEAMS.  Ql.  IV. 

When  experiments  on  beams  are  carried  to  the  point  of  rup- 
ture and  the  longitudinal  unit-stress  6"  computed  from  formula 
(4)  a  disagreement  of  that  value  with  those  found  by  direct 
experiments  on  tension  or  compression  is  observed.  This  is 
often  regarded  as  an  objection  to  the  common  theory  of  flexure, 
but  it  is  in  reality  no  objection,  since  law  (G)  and  formula  (4) 
are  only  true  provided  the  elastic  limit  of  the  material  be  not 
exceeded.  Experiments  on  the  deflection  of  beams  furnish  on 
the  other  hand  the  most  satisfactory  confirmation  of  the  theory. 
When  E  is  known  by  tensile  or  compressive  tests  the  formulas 
for  deflection  are  found  to  give  values  closely  agreeing  with 
those  observed.  Indeed  so  reliable  are  these  formulas  that  it 
is  not  uncommon  to  use  them  for  the  purpose  of  computing  E 
from  experiments  on  beams.  If  however  the  elastic  limit  of 
the  material  be  exceeded,  the  computed  and  observed  deflec- 
tions fail  to  agree. 

On  the  whole  it  may  be  concluded  that  the  common  theory 
of  flexure  is  entirely  satisfactory  and  sufficient  for  the  investi- 
gation of  all  practical  questions  relating  to  the  strength  and 
stiffness  of  beams.  The  actual  distribution  of  the  internal 
stresses  is  however  a  matter  of  very  much  interest  and  this  will 
be  discussed  at  some  length  in  Chapter  VIII. 

The  theory  of  flexure  is  here  applied  to  continuous  beams 
only  for  the  case  of  uniform  loads.  It  should  be  said  however 
that  there  is  no  difficulty  in  extending  it  to  the  case  of  concen- 
trated loads.  By  a  course  of  reasoning  similar  to  that  of  Art. 
48  it  may  be  shown  that  the  theorem  of  three  moments  for 
single  loads  is, 

M'l'  +  2M"(l'  +  /")  +  M'"l"  =  -  P'l'\k  -  &) 


Here  as  in  Fig.  37  the  moments  at  three  consecutive  supports 
are  designated  by  M',  M"  ,  and  M"'  and  the  lengths  of  the  two 
spans  by  /'  and  I".  P'  is  any  load  on  the  first  span  at  a  dis- 


ART.  50.      REMARKS   ON   THE  THEORY   OF  FLEXURE.  IOQ 

tance  kl'  from  the  left  support  and  P"  any  load  on  the  second 
span  at  a  distance  kl"  from  the  left  support,  k  being  any  frac- 
tion less  than  unity  and  not  necessarily  the  same  in  the  two 
cases.  From  this  theorem  the  negative  bending  moments  at 
the  supports  for  any  concentrated  loads  may  be  found,  and  the 
beam  be  then  investigated  by  formulas  (6)  and  (4).  For  ex- 
ample, if  a  beam  of  three  equal  spans  be  loaded  with  P  at  the 
middle  of  each  span,  the  negative  moments  at  the  supports 

are  each  $-Pl. 
20 

The  Journal  of  the  Franklin  Institute  for  March  and  April, 
1875,  contains  an  article  by  the  author  in  which  the  law  of  in- 
crease of  the  quantities  in  the  tables  of  Art.  48  is  explained 
and  demonstrated.  A  general  abbreviated  method  of  deduc- 
ing the  moments  at  the  supports  for  both  uniform  and  concen- 
trated loads  on  restrained  and  continuous  beams  is  given  in  the 
Philosophical  Magazine  for  September,  1875.  See  Text-book 
on  Roofs  and  Bridges,  Part  IV. 

Exercise  4.  Consult  BARLOW'S  Strength  of  Materials  (Lon- 
don, 1837),  and  write  an  essay  concerning  his  experiments  to 
determine  the  laws  of  the  strength  and  stiffness  of  beams.  Con- 
sult also  BALL'S  Experimental  Mechanics. 

Exercise  5.  Consult  Engineering  News,  Vol.  XVIII,  pp.309, 
352,  404,  443;  Vol.  XIX,  pp.  11,  28,  48,  84;  and  Vol.  XXII, 
p.  121.  Write  an  essay  concerning  certain  erroneous  views  re- 
garding the  theory  of  flexure  which  are  there  discussed.  Con- 
sult also  TODHUNTER'S  History  of  the  Elasticity  and  Strength 
of  Materials. 

Exercise  6.  Procure  six  sticks  of  ash  each  -J  X  f  inches  and 
of  lengths  about  8,  12,  and  16  inches.  Devise  and  conduct 
experiments  to  test  the  following  laws :  First,  the  strength  of  a 
beam  varies  directly  as  its  breadth  and  directly  as  the  square 
of  its  depth.  Second,  the  stiffness  of  a  beam  is  directly  as  its 
breadth  and  directly  as  the  cube  of  its  depth.  Third,  a  beam 
fixed  at  the  ends  is  twice  as  strong  and  four  times  as  stiff  as  a 


110  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

simple  beam  when  loaded  at  the  middle.  Write  a  report 
describing  and  discussing  the  experiments. 

Exercise  7.  In  order  to  test  the  theory  of  continuous  beams 
discuss  the  following  experiments  by  FRANCIS  and  ascertain 
whether  or  not  the  ratio  of  the  two  observed  deflections  agrees 
with  theory.  "A  frame  was  erected,  giving  4  bearings  in  the 
same  horizontal  plane,  4  feet  apart,  making  3  equal  spans,  each 
bearing  being  furnished  with  a  knife  edge  on  which  the  beam 
was  supported.  Immediately  over  the  bearings  and  secured  to 
the  same  frame  was  fixed  a  straight  edge,  from  which  the  de- 
flections were  measured.  A  bar  of  common  English  refined 
iron,  12  feet  2j  inches  long,  mean  width  1.535  inches,  mean 
depth  0.367  inches,  was  laid  on  the  4  bearings,  and  loaded  at 
the  center  of  each  span  so  as  to  make  the  deflections  the  same, 
the  weight  at  the  middle  span  being  82.84  pounds  and  at  each 
of  the  end  spans  52.00  pounds.  The  deflections  with  these 
weights  were, 

At  the  center  of  the  middle  span  0.281  inches. 

At  center  of  end  spans,  0.275  and  0.284  inches, 

mean  0.280  inches. 

A  piece  3  feet  uf  inches  long  was  then  cut  from  each  end  of 
the  bar,  leaving  a  bar  4  feet  4f  inches  long,  which  was  replaced 
in  its  former  position  and  loaded  with  the  same  weight  (82.84 
pounds)  as  before,  when  its  deflection  was  found  to  be  1.059 
inches." 

Prob.  86.  A  beam  of  three  spans,  the  center  one  being  /  and 
the  side  ones  nl,  is  loaded  with  P  at  the  middle  of  each  span. 
Find  the  value  of  n  so  that  the  reactions  at  the  end  may  be 
one-fourth  of  the  other  reactions. 

Prob.  87.  Let  a  beam  whose  cross-section  is  an  isosceles  tri- 
angle have  the  base  b  and  the  depth  d.  Prove  that  if  0.13^  be 
cut  off  from  the  vertex  the  remaining  trapezoidal  beam  will 
be  about  9  per  cent  stronger  than  the  triangular  one. 


ART.  51.  CROSS-SECTIONS   OF   COLUMNS.  Ill 


CHAPTER  V. 
COLUMNS   OR   STRUTS. 

ART.  51.    CROSS-SECTIONS  OF  COLUMNS. 

A  column  is  a  prism,  greater  in  length  than  about  ten  times 
its  least  diameter,  which  is  subject  to  compression.  If  the 
prism  be  only  about  four  or  six  times  as  long  as  its  least  diam- 
eter the  case  is  one  of  simple  compression,  the  constants  for 
which  are  given  in  Art.  6.  In  a  case  of  simple  compression 
failure  occurs  by  the  crushing  and  splintering  of  the  material, 
or  by  shearing  in  directions  oblique  to  the  length.  In  the  case 
of  a  column,  however,  failure  is  apt  to  occur  by  a  sidewise 
bending  which  causes  flexural  stresses.  The  word  *  strut '  is 
frequently  used  as  synonymous  with  column,  and  sometimes 
also  the  word  '  pillar.' 

Wooden  columns  are  usually  square  or  round  and  they  may 
be  built  hollow.  Cast  iron  columns  are  usually  round  and  they 
are  often  cast  hollow.  Wrought  iron  columns  are  made  of  a 
great  variety  of  forms.  I  beams  may  be  used,  but  most  columns 
are  usually  made  of  three  or  more  different  shape-irons  riveted 
together.  The  Phcenix  column  is  made  by  riveting  together 
flanged  circular  segments  so  as  to  form  a  closed  cylinder.  It  is 
clear  that  a  square  or  round  section  is  preferable  to  an  unsym- 
metrical  one,  since  then  the  liability  to  bending  is  the  same  in 
all  directions.  For  a  rectangular  section  the  plane  of  flexure 
will  evidently  be  perpendicular  to  the  longer  side  of  the  cross- 
section,  and  in  general  the  plane  of  flexure  will  be  perpendicular 
to  that  axis  of  the  cross-section  for  which  the  moment  of  in- 
ertia is  the  least.  In  designing  a  column  it  is  hence  advisable 


112 


COLUMNS   OR   STRUTS. 


CH.  V. 


that  the  cross-section  should  be  so  arranged  that  the  moments 
of  inertia  about  the  two  principal  rectangular  axes  may  be 
approximately  equal. 

For  instance,  let  it  be  required  to  construct  a  column  with 
two  I  shapes  and  two  plates  as  shown  in  Fig.  43.  The  I  beams 
are  to  be  light  lo-inch  ones  weighing  30 
pounds  per  linear  foot,  and  having  the 
flanges  4.32  inches  wide.  The  plates  are 
to  be  £  inch  thick,  and  it  is  required  to 
find  their  length  x  so  that  the  liability  to 
bending  about  the  two  axes  shown  in  the 
figure  may  be  the  same.  From  the  manu- 
facturer's  table  it  is  found  that  the  moment 


43- 


of  inertia  /  of  the  beam  about  an  axis  through  its  center  of 
gravity  and  perpendicular  to  the  web  is  150,  while  the  moment 
of  inertia  I'  about  an  axis  through  the  same  point  and  parallel 
to  the  web  is  nearly  8.  Hence,  for  the  axes  showji  in  the 
figure,  the  moments  of  inertia  are, 
For  axis  perpendicular  to  plates, 

/•»       *  3      /^  I          _ .     v    >      O        I         ,• .     v    > 


12 


For  axis  parallel  to  plates, 
„*  X  0.5" 


12 


+  2  X  0.5*  x  5-25'  +  2  X  150. 


Placing  these  two  expressions  equal,  the  value  of  x  is  found  to 
be  between  14  and  14^  inches. 

Prob.  88.  A  column  is  to  be  formed  of  two  light  1 2-inch  eye- 
beams  connected  by  a  lattice  bracing.  Find  the  proper  distance 
between  their  centers,  disregarding  the  moment  of  inertia  of 
the  latticing. 

Prob.  89.  Two  joists  each  2X4  inches  are  to  be  placed  6 
inghes  apart  between  their  centers,  and  connected  by  two  others 
each  8  inches  wide  and  x  inches  thick  so  as  to  form  a  closed 
hollow  rectangular  column.  Find  the  proper  value  of  x. 


ART.  52. 


GENERAL   PRINCIPLES. 


ART.  52.    GENERAL  PRINCIPLES. 

If  a  short  prism  of  cross-section  A  be  loaded  with  the  weight 
P,  the  internal  stress  is  to  be  regarded  as  uniformly  distributed 

over  the  cross-section,  and  hence  the  compressive  unit-stress  Se 

p 
is  --.     But  for  a  long  prism,  or  column,  this  is  not  the  case; 

A 

P 
while  the  average  unit-stress  is  — r ,  the  stress  in  certain  parts  of 

the  cross-section  may  be  greater  and  upon  others  less  than  this 
value  on  account  of  the  transverse  stresses  due  to  the  sidewise 
flexure.  Hence  in  designing  a  column  the  load  P  must  be 
taken  as  smaller  for  a  long  one  than  for  a  short  one,  since  evi- 
dently the  liability  to  bending  increases  with  the  length. 

Numerous  tests  on  the  rupture  of  long  columns  have 
shown  that  the  load  causing  the  rupture  is  approximately  in- 
versely proportional  to  the  square  of  the  length  of  the  column. 
That  is  to"  say,  if  there  be  two  columns  of  the  same  material 
and  cross-section  and  one  twice  as  long  as  the  other,  the  long 
one  will  rupture  under  about  one-quarter  the  load  of  the 
short  one. 

The  condition  of  the  ends  of  columns  exerts  a  great  influence 
upon  their  strength.    Class  (a)  includes  those  with  '  round  ends/ 
or  those  in  such  condition  that  they  are  free  to  turn  at  the  ends. 
Class  (c)  includes  those  whose  ends 
are  '  fixed '  or  in  such  condition  that 
the  tangent  to  the  curve  at  the  ends 
always   remains   vertical.     Class  (b) 
includes   those  with    one  end  fixed 
and  the  other  round.    In  architecture 
it  is  rare  that  any  other  than  class  (c) 
is  used.     In  bridge  construction  and 
in  machines,  however,    columns    of 
classes  (b)  and  (a)  are  very  common.     It  is  evident  that  class  (c) 


114  COLUMNS  OR   STRUTS.  CH.  V. 

is  stronger  than  (£),  and  that  (b)  is  stronger  than  (a),  and  this  is 
confirmed  by  all  experiments.  Fig.  44  is  intended  as  a  sym- 
bolical representation  of  the  three  classes  of  columns,  and  not 
as  showing  how  the  ends  are  rendered  'round'  and  'fixed'  in 
practical  constructions. 

The  theory  of  the  resistance  of  columns  has  not  yet  been 
perfected  like  that  of  beams,  and  accordingly  the  formulas  for 
practical  use  are  largely  of  an  empirical  character.  The  form 
of  the  formulas  however  is  generally  determined  from  certain 
theoretical  considerations,  and  these  will  be  presented  in  the 
following  articles  as  a  basis  for  deducing  the  practical  rules. 

Prob.  90.  A  column  of  \\rouHit-iron  shapes  weighing  186 
pounds  per  yard  is  1  1  inches  square  and  3  feet  long.  Whc*c 
load  will  it  carry  with  a  factor  of  safety  of  5  ? 

ART.  53.    EULER'S  FORMULA  FOR  LONG  COLUMNS. 

Consider  a  column  of  cross-section  A  loaded  with  a  weight  P 
|p         under  whose  action  a  certain  small  sidewise  bending 
4          occurs.     Let  the  column  be  round,  or  free  to  turn  at 
both  ends  as  in  Fig.  45.     Take  the  origin  at  the  upper 
end,  and  let  x  be  the  vertical  and  y  the  horizontal  co- 
ordinate of  any  point  of  the  elastic  curve.     The  gen- 
eral equation  (5)  deduced   in  Art.    33,  applies   to   all 
bodies   subject   to    flexure    provided   the   bending   be 
slight  and  the  elastic  limit  of  the  material  be  not  ex- 
ceeded.     For   the   column   the   bending    moment   is 
Fig.  45.     —  Py,  the  negative  sign  being  used  because  the  curve 
is  concave  to  the  axis  of  x  ;  hence, 


.   0 

The  first  integration  of  this  gives,    (  ^ 


ART.  53. 


EULER'S  FORMULA. 


dy 
But  when  y  =  the  maximum  deflection  ^,  the  tangent  ~-  =  o. 

Hence  C  =  P4*,  and  by  inversion, 

dx  =  (^^=, 

The  second  integration  now  gives, 


=(?)' 


arc  sin  ~  4-  £T'. 

' 


Here  C"'  is  o  because  /  =  o  when  x  =  o.     Hence  finally  the 
equation  of  the  elastic  curve  of  the  column  is, 


This  equation  is  that  of  a  sinusoid.     But  also  y  =  o  when 

/  P\* 
x  =  I.     Hence  if  n  be  an  integer,  IVJ^T]   must  equal  nn,  or, 

p  •=£&-. 


which  is  EULER'S  formula  for  the  resistance  of  columns.     This 
reduces  the  equation  of  the  sinusoid  to, 

y  =  A  sin  nn-j. 

The  three  curves  for  n  =  I,  n  =  2,  and  n  —  3  are  shown  in 
Fig.  46.  In  the  first  case  the 
curve  is  entirely  on  one  side  of 
the  axis  of  x,  in  the  second  case 
it  crosses  that  axis  at  the  mid- 
dle, and  in  the  third  case  it 
crosses  at  \l  and  \ /,  the  points 
of  crossing  being  also  inflection 
points  where  the  bending  mo- 
ment is  zero.  Evidently  the 


Fig.  46. 

greatest  deflection  will  occur  for   the  case  where  n  =  i,  and 


. 


1  16  COLUMNS   OR   STRUTS.  CH.  V. 

this  is  the  most  dangerous  case.     Hence, 

n*EI 
(a)  p=—jr-) 

is  EULER'S  formula  for  long  columns  with  round  ends. 

A  column  with  one  end  fixed  and  the  other  round  is  closely 
represented  by  the  portion  b'b"  of  the  second  case,  b'  being 
the  fixed  end  where  the  tangent  to  the  curve  is  vertical.  Here 
n  =  2,  and  the  length  b'b"  is  three-fourths  of  the  entire  length, 
hence, 


is  RULER'S  formula  for  long  columns  with  one  end  fixed  and 
the  other  round. 

A  column  with  fixed  ends  is  represented  by  the  portion  c'c" 
of  the  case  c.  Here  n  =  3,  and  the  length  c'c"  is  two-thirds  of 
the  entire  length,  hence, 


... 

is  EULER'S  formula  for  long  columns  with  fixed  ends. 

From  this  investigation  it  appears  that  the  relative  resist- 
ances of  long  columns  of  the  classes  (a),  (£),  and  (c)  are  as  the 
numbers  i,  2j,  and  4,  when  the  lengths  are  the  same,  and  this 
conclusion  is  approximately  verified  by  experiments.  It  also 
appears  that,  if  the  resistance  of  three  columns  of  the  classes 
(a),  (b),  and  (V)  are  to  be  equal,  their  lengths  must  be  as  the 
numbers  I,  if,  and  2. 

The  moment  of  inertia  7  in  the  above  formulas  is  taken  about 
a  neutral  axis  of  the  cross-section  perpendicular  to  the  plane  of 
the  flexure,  and  in  general  is  the  least  moment  of  inertia  of 
that  cross-section,  since  the  column  will  bend  in  the  direction 
which  offers  the  least  resistance.  If  A  be  the  area  of  the 


ART.  54.  HODGKINSON'S  FORMULAS.  117 

cross-section  and  r  its  least  radius  of  gyration,  the  value  of  / 
is  Ar*t  and  EULER's  formula  may  be  written 

p         y 


where  m  =  i,  2j,  or  4,  for  the  three  end  conditions. 

The  value  of  P  in  EULER'S  formula  gives  the  load  which 
holds  the  column  in  equilibrium  when  it  is  laterally  deflected. 
If  the  load  be  less  than  this  value  of  P  the  column,  if 
deflected,  will  return  to  its  original  straight  position.  If  the 
load  be  slightly  greater  than  P  the  bending  increases  until 
failure  occurs.  EULER'S  formula  is  hence  the  criterion  of 
indifferent  equilibrium  or  the  condition  for  the  failure  of  a 
column  by  lateral  bending.  The  maximum  deflection  A  is 
indeterminate,  since  it  cannot  be  found  from  the  equation  of 
the  elastic  curve. 

EULER'S  formula  is  little  used  in  practice,  except  in  Ger- 
many. When  so  used  it  takes  the  form 

P  _  rnrfE    S 

A~     ~7~'T" 

in  which  f  is  an  assumed  factor  of  security  and  P  is  the  load 
on  the  column. 

Prob.  91.  Show  that  the  value  of  the  constant  m  for  a  long 
column  which  is  fixed  at  one  end  and  entirely  free  at  the  other 
is  m  =  J. 

ART.  54.     HODGKINSON'S  FORMULAS. 

EULER'S  formula  gives  valuable  information  regard- 
ing the  laws  of  flexure  of  long  columns.  For  cylindrical 
columns  it  shows,  since  /=  Ttd'/G^.,  that  the  load  P  which 
causes  lateral  failure  varies  directly  as  the  fourth  power  of 
the  diameter  and  inversely  as  the  square  of  the  length. 
HODGKINSON  in  his  experiments  observed  that  this  was 
approximately  but  not  exactly  the  case.  He  therefore 


Il8  COLUMNS   OR   STRUTS.  Cli.  V. 

wrote  for  each  kind  of  columns  the  analogous  expression, 


and  determined  the  constants  a,  /?,  and  6  from  the  results  of 
his  experiments,  thus  producing  empirical  formulas. 

Let  P  be  the  crushing  load  in  gross  tons,  d  the  diameter  of 
the  column  in  inches,  and  /  its  length  in  feet.  Then  HODG- 
KINSON'S  empirical  formulas  are, 

For  solid  cast  iron  cylindrical  columns, 

P  =  14.9—-         for  round  ends, 


for  flat  ends, 


For  solid  wrought  iron  cylindrical  columns, 

^3-76 
P  =  42-7^—          for  round  ends; 

^3.76 
P  =  134-—         for  flat  ends. 

These  formulas  indicate  that  the  ultimate  strength  of  flat-ended 
columns  is  about  three  times  that  of  round-ended  ones.  The 
experiments  also  showed  that  the  strength  of  a  column  with 
one  end  flat  and  the  other  end  round  is  about  twice  that  of 
one  having  both  ends  round.  HODGKINSON'S  tests  were  made 
upon  small  columns  and  his  formulas  are  not  so  reliable  as 
those  which  will  be  given  in  the  following  articles.  For  small 
cast  iron  columns  however  the  formulas  are  still  valuable. 

By  the  help  of  logarithms  it  is  easy  to  apply  these  formulas 
to  the  discussion  of  given  cases.  Usually  P  will  be  given  and 
d  required,  or  d  be  given  and  P  required.  By  using  assumed 
factors  of  safety  the  proper  size  of  cylindrical  columns  to  carry 
given  loads  may  also  be  determined.  These  formulas,  it  should 
be  remembered,  do  not  apply  to  columns  shorter  than  about 


ART.  55.  RANKINE'S  FORMULA.  119 

thirty  times  their  least  diameters.     The  word  flat  used  in  this 
Article  is  to  be  regarded  as  equivalent  to  fixed. 

Prob.  92.  A  cast-iron  cylindrical  column  with  flat  ends  is  3 
inches  diameter  and  8  feet  long.  What  load  will  cause  it 
to  fail? 

Prob.  93.  A  cast-iron  cylindrical  column  with  flat  ends  is  to 
be  7  feet  long  and  carry  a  load  of  200  ooo  pounds  with  a  factor 
of  safety  of  6.  Find  the  proper  diameter. 


ART.  55.    RANKINE'S  FORMULA. 

The  columns  which  are  generally  employed  in  engineering 
practice  are  intermediate  in  length  between  short  prisms  and 
the  long  columns  to  which  EULER'S  formula  ap- 
plies.    They  fail  under  the   stresses   caused   by 
combined    flexure    and    compression.      Fig.    47 
shows  the  flexure  very  much  exaggerated.     The 

P 

load  P  produces  an  average  unit-stress  —  on  any 

A 

horizontal  cross-section  whose  area  is  A,  but  in 
consequence  of  the  bending  this  is  increased  on 
the  concave  side  and  diminished  on  the  convex 
side  by  an  amount  S^  The  value  of  Sl  depends 
upon  the  bending  moment  PA,  where  A  is  the 
lateral  deflection  at  the  middle  of  the  column. 

The  total  unit-stress  on  the  concave  side  is  then 

p  Fig.  47. 

—  +  Slt  and  it  isnatural  to  suppose  that  failure 

A. 

will  occur  when  this  is  equal  to  the  ultimate  strength  of  the 
material.  Many  formulas  have  been  proposed  for  the  investi- 
gation of  such  columns,  but  all  of  them  are  more  or  less  em- 
pirical, as  certain  constants  are  derived  from  the  results  of  ex- 


120  COLUMNS  OR  STRUTS.  CH.  V. 

periments  upon   the  rupture  of  columns,  or  assumptions  are 
made  regarding  the  form  of  the  formula. 

The  formulas  of  EULER  are  unsuitable  for  practical  cases  of 
investigation  because  they  contain  no  constant  indicating  the 
working  or  ultimate  compressive  strength  of  the  material,  and 
because  they  apply  only  to  long  columns.  HODGKINSON's 
formulas  are  unsatisfactory  for  similar  reasons,  and  because 
they  do  not  well  agree  with  later  experiments. 

The  formula  which  appears  to  have  the  best  theoretical 
foundation  will  now  be  presented.  It  is  sometimes  called 
GORDON'S  formula,  and  occasionally  it  is  referred  to  as  "  GOR- 
DON'S formula  modified  by  RANKINE,"  but  the  best  usage 
gives  to  it  the  name  of  RANKINE'S  formula.  (The  formula 
deduced  by  GORDON  differs  from  (10)  in  that  it  applies  only  to 
rectangular  or  circular  cross-sections,  r  being  replaced  by  d,  the 
least  side  or  diameter,  and  q  having  different  values  from  those 
given  in  the  table.) 

Let  P  be  the  load  on  the  column,  /  its  length,  A  the  area  of 
its  cross-section,  /  the  moment  of  inertia,  and  r  the  radius  of 
gyration  of  that  cross-section  with  reference  to  a  neutral  axis 
perpendicular  to  the  plane  of  flexure,  and  c  the  shortest  dis- 
tance from  that  axis  to  the  remotest  fiber  on  the  concave  side. 
The  average  compressive  unit-stress  on 

p 

any  cross-section  is  — ,  but  in  consequence 
A 

of  the  flexure  this  js  increased  on  the 
concave  side,  and  decreased  on  the  con- 
vex  side.  Thus  in  Fig.  48  the  average 

p 

unit-stress  — -  is  represented  by  cd,  but  on 

the  concave  side  this  is  increased  to  aq,  and  on  the  convex 
side  decreased  to  bq.      The  triangles  pdq  and    qdp  represent 


ART.  55-  RANKINE'S  FORMULA.  121 

the  effect  of  the  flexure  exactly  as  in  the  case  of  beams,  pq 
indicating  the  greatest  compressive  and  qp  the  greatest  ten- 
sile unit-stress  due  to  the  bending.  Let  the  total  maximum 
unit-stress  aq  be  denoted  by  5  and  the  part  due  to  the  flexure 
be  denoted  by  S19  Then, 


New,  from  the  fundamental  formula  (4)  the  flexural  stress  is 

Me 

—  ,  where  M  is  the  external  bending  moment,  which  for  a  col- 

umn has  its  greatest  value  when  M  =  PA,  A  being  the  maxi- 
mum deflection.  /  =  Ar*  is  the  well-known  relation  between  / 
and  r.  Hence  the  value  of  Sl  is, 

PAc      PAc 


By  analogy  with  the  theory  of  beams,  as  in  Art.  37,  the  value 

^» 
of  A  may  be  regarded  as  varying  directly  as  —  .     Hence  if  q  be 

a  quantity  depending  upon  the  kind  of  material  and  the  con- 
dition of  the  ends,  the  total  unit-stress  is, 

P 


This  may  now  be  written  in  the  usual  form, 

P  _        S 
(10)  A  =         —» 


which  is  RANKINE'S  formula  for  the  investigation  of  columns. 

The  above  reasoning  has  been  without  reference  to  the  ar- 
rangement of  the  ends  of  the  column.  By  Art.  53  it  is  known 
that  a  column  with  round  ends  must  be  one  half  the  length  of 
one  with  fixed  ends  in  order  to  be  of  equal  strength,  and  that 
a  column  with  one  end  fixed  and  the  other  round  must  be 


122 


COLUMNS   OR   STRUTS. 


CH.  V. 


three  fourths  the  length  of  one  with  fixed  ends  in  order  to  be 
of  equal  strength.  Therefore  if  q  be  the  constant  for  fixed 
ends,  (£fq  will  be  the  constant  for  one  end  fixed  and  the  other 
round,  and  2*q  will  be  the  constant  for  both  ends  round. 

The  values  of  q  to  be  taken  for  use  in  formula  (10)  for  the 
examples  and  problems  of  this  chapter  may  be  the  following 
rough  values,  unless  otherwise  stated,  while  the  values  of 
the  ultimate  compressive  unit-stress  6"  will  be  taken  from 
the  table  in  Art.  6. 


Material. 

Both  Ends 
Fixed. 

Fixed  and 
Round. 

Both    Ends 
Round. 

Timber, 
Cast  Iron, 
Wrought  Iron, 
Steel, 

I 

I.78 

4 

3  ooo 

i 

3  ooo 
1.78 

3  ooo 

4 

5  ooo 

i 

5  ooo 
i.78 
36  ooo 

1.78 

5  ooo 
4 

36  ooo 
I 

36000 
4 

25  ooo 

25  ooo 

25  ooo 

Very  wide  variation  in  the  values  of  q  is  found  from  experi- 
ments on  the  rupture  of  different  types  of  columns.  Formula 
(10)  when  used  with  such  experimental  values  is  an  empirical 
one  only.  In  Art.  61  it  is  shown  how  a  theoretic  value  of  q 
may  be  derived  which  renders  (10)  a  rational  formula. 

Prob.  94.   Plot  the   curve  represented  by  formula  (10)  for 
cases  of  wrought-iron  columns  with  fixed  and  with  round  ends, 

P  I 

taking  the  values  of  —  as  ordinates,  and  the  values  of  —  as  ab- 

jrL  T 

scissas. 

ART.  56.    RADIUS  OF  GYRATION  OF  CROSS-SECTIONS. 

The  radius  of  gyration  of  a  surface  with  reference  to  an  axis 
is  equal  to  the  square  root  of  the  ratio  of  the  moment  of  iner- 


ART.   57.  INVESTIGATION  OF  COLUMNS.  123 

tia  of  the  surface  referred  to  the  same  axis  to  the  area  of  the 
figure.  Or  if  r  be  radius  of  gyration,  /  the  moment  of  inertia, 
and  A  the  area  of  the  surface,  then  I =Ar*. 

In  the  investigation  of  columns  by  formula  (10)  the  value  of 
ra  is  required,  r  being  the  least  radius  of  gyration.  These 
values  are  readily  derived  from  the  expressions  for  the  mo- 
ment of  inertia  given  in  Art.  23,  the  most  common  cases  being 
the  following, 

For  a  rectangle  whose  least  side  is  d,         r*  =  — . 

For  a  circle  of  diameter  d,  r*  =  -^. 

16 

For  a  triangle  whose  least  altitude  is  d,      r*  =  —$. 

Io 

For  a  hollow  square  section,  r1  = 

For  a  hollow  circular  section,  r*  = 


12 

d'+d" 


16 

For  I  beams  and  other  shapes,  r9  is  found  by  dividing  the  least 
moment  of  inertia  of  the  cross-section  by  the  area  of  that  cross- 
section.  For  instance,  by  the  help  of  the  table  in  Art.  30,  the 
least  value  of  ra  for  a  light  12-inch  I  beam  is  found  to  be 
g=  1.02  inches'. 

Prob.  95.  Compute  the  least  radius  of  gyration  for  a  T  iron 
whose  width  is  4  inches,  depth  4  inches,  thickness  of  flange  J 
inches,  and  thickness  of  stem  £f  inches. 

ART.  57.  INVESTIGATION  OF  COLUMNS. 
The  investigation  of  a  column  consists  in  determining  the 
maximum  compressive  unit-stress  5"  from  formula  (10).  The 
values  of  P,  A,  /,  and  r  will  be  known  from  the  data  of  the  given 
case,  and  q  is  known  from  the  results  of  previous  experiments. 
Then,  p 


124  COLUMNS  OR  STRUTS.  CH.  V. 

and,  by  comparing  the  computed  value  of  5  with  the  ultimate 
strength  and  elastic  limit  of  the  material,  the  factor  of  safety 
and  the  degree  of  stability  of  the  column  may  be  inferred. 

For  example,  consider  a  hollow  wooden  column  of  rectangu- 
lar section,  the  outside  dimensions  being  4X5  inches  and  the 
inside  dimensions  3X4  inches.  Let  the  length  be  18  feet, 
the  ends  fixed,  and  the  load 'be  5400  pounds.  Here  P  = 
5  400,  A  —S  square  inches,  !  =  216  inches.  From  the  table 
q  =  3-^.  From  Art.  56, 

- -  **£*¥* = -•    

Then  the  substitution  of  these  values  gives, 

5400/  2r6X2i6\ 

5  =  — g—  (i  +  3000X22J  =  5430  pounds  per  square  inch. 

Here  the  average  unit-stress  is  675  pounds  per  square  inch, 
but  the  flexure  has  increased  that  stress  on  the  concave  side  to 
5  430  pounds  per  square  inch,  so  that  the  factor  of  safety  is 
only  about  i£. 

Prob.  96.  A  cylindrical  wrought  iron  column  with  fixed  ends 
is  12  feet  long,  6.36  inches  in  exterior  diameter,  6.02  inches  in 
interior  diameter,  and  carries  a  load  of  98  ooo  pounds.  Find 
its  factor  of  safety. 

Prob.  97.  A  pine  stick  3X4  inches  and  12  feet  long  is  used 
as  a  column  with  fixed  ends.  Find  its  factor  of  safety  under 
a  load  of  4  ooo  pounds.  If  the  length  be  only  one  foot,  what  is 
the  factor  of  safety? 

ART.  58.    SAFE  LOADS  FOR  COLUMNS. 

To  determine  the  safe  load  for  a  given  column  it  is  necessary 
to  first  assume  the  allowable  working  unit-stress  5.  Then  from 
formula  (10)  the  safe  load  is, 


r 


ART.  59.  DESIGNING   OF   COLUMNS.  12$ 

Here  A,  /,  and  r  are  known  from  the  data  of  the  given  problem 
and  q  is  taken  from  the  table  in  Art.  55. 

For  example,  let  it  be  required  to  determine  the  safe  load 
for  a  fixed-ended  timber  column,  3X4  inches  in  size  and  10 
feet  long,  so  that  the  greatest  compressive  unit-stress  may  be 
800  pounds  per  square  inch.  From  the  formula, 

800  X  12 

P  =  -  — = =  about  i  300  pounds. 

120' 

3  ooo  X  | 

A  short  prism   3X4  inches  should  safely  carry  seven  times 
this  load. 

Prob.  98.  Find  the  safe  load  for  a  heavy  steel  I  beam  of 
15  inches  depth  and  10  feet  length  when  used  as  a  column  with 
fixed  ends,  the  factor  of  safety  being  4. 

Prob.  99.  Find  the  safe  steady  load  for  a  hollow  cast  iron 
column  with  fixed  ends,  the  length  being  18  feet,  outside 
dimensions  4X5  inches,  inside  dimensions  3X4  inches. 


ART.  59.    DESIGNING  OF  COLUMNS. 

When  a  column  is  to-  be  selected  or  designed  the  load  to  be 
borne  will  be  known,  as  also  its  length  and  the  condition  of  the 
ends.  A  proper  allowable  unit-stress  5  is  assumed,  suitable  for 
the  given  material  under  the  conditions  in  which  it  is  used. 
Then  from  formula  (i)  the  cross-section  of  a  short  column  or 

P 

prism  is  -=-,  and  it  is  certain  that  a  greater  value  of  the  cross- 
section  than  this  will  be  required.  Next  assume  a  form  and 
area  A,  find  r2,  and  from  the  formula  (10)  compute  5".  If  the 
computed  value  agrees  with  the  assumed  value  the  correct  size 
has  been  selected.  If  not,  assume  a  new  area  and  compute  .S 
again,  and  continue  the  process  until  a  proper  agreement  is 
attained. 


126  COLUMNS  OR  STRUTS.  CH.  V. 

For  example,  a  hollow  cast  iron  rectangular  column  of  18  feet 
length  is  to  carry  a  load  of  60  ooo  pounds.  Let  the  working 
strength  .S  be  15  OOO  pounds  per  square  inch.  Then  for  a  short 
length  the  area  required  would  be  four  square  inches.  Assume 
then  that  about  6  square  inches  will  be  needed.  Let  the  sec- 
tion be  square,  the  exterior  dimensions  6x6  inches,  and  the  in- 
terior dimensions  5^  X  5i  inches.  Then  A  =  5.75,  1=  18  X  12, 
P—  60000,  q—  ^5,  r2  —  5.52,  and  from  (10), 

5  = 


6oooo/  i82  X  12'    \ 

--  1  1  +  --  j  =  about  30  ooo, 
5.75    \     [   5000  X  5-52' 


which  shows  that  the  dimensions  are  much  too  small.  Again 
assume  the  exterior  side  as  6  inches  and  the  interior  as  5  inches. 
Then  A  =  n,  r2  =  5.08,  and 

6oooo/  i82  X  12" 


As  this  is  very  near  the  required  working  stress,  it  appears  that 
these  dimensions  very  nearly  satisfy  the  imposed  conditions. 

In  many  instances  it  is  possible  to  assume  all  the  dimensions 
of  the  column  except  one,  and  then  after  expressing^  and  r  in 
terms  of  this  unknown  quantity,  to  introduce  them  into  (10)  and 
solve  the  problem  by  finding  the  root  of  the  equation  thus 
formed.  For  example,  let  it  be  required  to  find  the  size  of  a 
square  wooden  column  with  fixed  ends  and  24  feet  long  to  sus- 
tain a  load  of  100000  pounds  with  a  factor  of  safety  of  10. 

Here  let  x  be  the  unknown  side ;  then  A  =  x*  and  r*  =  — . 

From  (10), 

100 ooo/          24"  X  i: 

800  = 5 —  i  H 

x       v         3  ooo  x 

By  reduction  this  becomes, 

Sx4  —  i  ooo*'  =  331  776, 
the  solution  of  which  gives  16.6  inches  for  the  side  of  the  column. 


ART.  60.  THE   STRAIGHT-LINE   FORMULA.  I2/ 

Prob.  ioo.  Find  the  size  of  a  square  wooden  column  with  fixed 
ends  and  12  feet  in  length  to  sustain  a  load  of  100000  pounds 
with  a  factor  of  safety  of  10.  Find  also  its  size  for  round  ends. 


ART.  60.    TrtE  STRAIGHT-LINE  FORMULA. 

In  1886  a  straight-line  formula  for  columns,  as  a  substitute 
for  RANKINE'S,  was  proposed  by  THOMAS  H.  JOHNSON,  which 
has  since  been  extensively  used  on  account  of  its  simplicity 
when  expressed  in  numerical  form.  The  notation  being  the 
same  as  in  Art.  55,  this  formula  is, 

p  -s     kl 

~A~  r' 

in  which  k  is  a  constant  whose  value  is, 

t       S 
'    3 


where  m  is  I,  2j,  or  4,  depending  on  the  condition  of  the  ends, 
as  in  EULER'S  formula  (Art.  53). 

This  formula  is  not  a  rational  one,  the  equation  of  the 
straight  line  being  assumed  merely  as  a  good  representation 
of  the  results  of  experiments  on  the  rupture  of  columns.  The 
value  of  k  is  deduced  by  making  the  straight  line  tangent  to  the 
curve  which  represents  EULER'S  formula.  Thus,  let  the  values  of 

P  I 

—  and  -  be  regarded  as  the  ordinates  and  abscissas  of  a  curve, 

/i  T 

and  be  designated  by  u  and  v  respectively.     Then  the  equa- 
tions of  EULER'S  curve  and  of  the  assumed  straight  line  are, 


~ 
u  —  -  5—         and         u  =  S  —  kv. 

By  placing  equal  the  values  of  u  in  these  two  equations  and 
also  the  values  of  the  first  derivatives,  the  ordinate  and  abscissa 
of  the  point  of  tangency  are  found  to  be, 

i  o 
*  =    S         and        z/ 


128 


COLUMNS   OR  STRUTS. 


CH.  V. 


and  then  the  value  of  k,  as  above  ghnen.  results.     The  value  of 
i\  is  the  limiting  value  of  -,  within  which  the  straight-line 

formula  is  to  be  used. 

The  values  of  5  to  be  used  for  cases  of  rupture  are  such  as 
make  the  straight-line  agree  best  with  experimental  results. 
The  values  derived  by  JOHNSON  in  his  discussion  are  given  in 
the-  following  table,  together  with  the  corresponding  values  of 

k  and  limiting  values  of  - ,  which  are  computed  by  taking  m  as 
2j,  if,  and  I  for  the  given  cases. 


Kind  of  Column. 

5 

k 

Limit  — 
r 

Wrought  iron: 

Flat  ends, 

42  ooo 

128 

218 

Hinged  ends, 

42  ooo 

157 

178 

Round  ends, 

42  ooo 

203 

I38 

Mild  steel: 

Flat  ends, 

52  500 

J79. 

195 

Hinged  ends, 

52  500 

220 

159 

Round  ends, 

52  500 

284 

123 

Cast  iron: 

Flat  ends, 

80  ooo 

438 

122 

Hinged  ends, 

80  ooo 

537 

99 

Round  ends, 

80  ooo 

693 

77 

Oak: 

Flat  ends, 

5400 

28 

128 

It  will  be  noticed  that  the  values  of  S  in  the  above  table  are 
less  than  the  average  values  of  ultimate  strength  given  in  Art. 
6.  For  ductile  materials,  like  wrought  iron  and  mild  steel,  this 
should  be  the  case  in  columns,  since  when  the  elastic  limit  is 
passed  a  flow  of  metal  begins  which  causes  the  lateral  deflec- 
tion to  increase,  and  failure  then  rapidly  follows. 

Reference  is  made  to  JOHNSON'S  paper  in  Transactions  of 


ART.  6l.  HITTER'S   RATIONAL  FORMULA.  129 

American  Society  of  Civil  Engineers  for  July  1886,  for  a 
fuller  discussion  of  the  straight-line  formula.  Although  much 
used  for  computing  values  of  P/A,  it  is  inconvenient  for  find- 
ing S,  since  this  quantity  is  included  in  k  and  a  cubic  equa- 
tion in  5  results. 

Prob.  101.   Solve  Problems  99  and  100  by  the  straight-line 
formula,  using  the  values  given  in  the  table. 

ART.  61.     RITTER'S  RATIONAL  FORMULA. 

Several  attempts  have  been  made  to  establish  a  formula  for 
columns,  which  shall  be  theoretically  correct,  like  formula  (4) 
for  beams,  when  the  material  is  not  stressed  beyond  the 
elastic  limit.  The  most  successful  attempt  is  that  of  RlTTER, 
who  in  1873  proposed  the  formula 
P  S 


in  which  Pis  the  load  on  the  column,  /its  length,  A  the  area 
and  r  the  least  radius  of  gyration  of  the  cross-section,  E  the 
coefficient  of  elasticity,  Se  the  unit-stress  at  the  elastic  limit, 
and  5  the  greatest  compressive  unit-stress  on  the  concave 
side,  while  m  =  4  when  both  ends  are  fixed,  m  =  2\  when 
one  end  is  round  and  the  other  fixed,  m  =  i  when  both  ends 
are  round,  and  m  =  i  when  one  end  is  fixed  and  the  other 
free. 

The  form  of  this  formula  is  the  same  as  that  of  RANKINE'S 
formula,  (10)  in  Art.  55,  but  it  deserves  a  special  name 
because  it  completes  the  deduction  of  the  latter  formula  by 
finding  for  q  a  value  which  is  closely  correct  when  the  stress 
5  does  not  exceed  the  elastic  limit  St. 

To  justify  RITTER'S  formula  let  it  be  noted  that  EULER'S 
formula  (Art.  53)  gives  the  value  of  P  which  causes  the  failure 
of  a  long  column  by  lateral  bending.  In  the  actual  long 
column,  however,  the  load  must  be  less  than  given  by 


130  COLUMNS  OR  STRUTS.  CH.  V 

EULER'S  formula,  since  it  is  required  that  stable  equilibrium 
shall  prevail.     Now,  if  there  be  written 
P  _    S  r* 

A~~se'mn    '7* 

for  the  long  column,  stable  equilibrium  prevails  when  S  is  less 
than  Se ,  while  failure  occurs  when  5  equals  Se ,  because  then 
the  column  will  not  spring  back  if  laterally  deflected  by  a 
slight  force.  Se/S  is  hence  the  factor  of  security /for  a  long 
stable  column,  as  noted  at  the  end  of  Art.  53.  Now  RlTTER'  I 
formula  reduces  to  the  above  form  for  long  columns  when  l/r 
is  so  great  that  unity  in  the  denominator  may  be  neglected  in 
comparison  with  the  following  term.  It  also  reduces  to  the 
form  P/A  =  5  for  short  blocks,  when  /  =  o.  The  formula 
hence  satisfies  the  two  limiting  conditions,  and  its  general  form 
is  justified  by  the  reasoning  of  Art.  55. 

Another  demonstration  may  be  given  as  follows:  Let  P/A 
be  denoted  by  y  and  l/r  by  x.  Then  the  formulas  of  RAN- 
KINE  and  of  EULER  for  stable  equilibrium  are 

5  mifE 

y  =  ,  .1.  ^  and  y  =  -yy-- 


Now  let  the  curves  which  these  equations  represent  be  tan- 
gent to  each  other.  For  the  point  of  tangency  the  two 
ordinates  are  equal,  as  also  the  two  derivatives  of  y  with 
respect  to  x.  Solving  these  two  equations  there  results 
x^  =  oo  for  the  point  of  tangency,  while 

Sf  S< 


mn*E  ~ 
is  the  constant  in  RANKINE'S  formula.     Thus, 

P 5 

A  -  Se      /" 


which  is  the  rational  formula  of  RITTER. 


ART.  61. 


HITTER'S  RATIONAL  FORMULA. 


To  find  the  mean  theoretical  values  of  q  the  mean  values 
of  Se  and  E,  as  stated  in  Art.  6,  may  be  used,  whence  results 
the  following  table: 

RATIONAL  VALUES  OF  q  FOR  FORMULA  (to). 


Material. 

Both  Ends 
Fixed. 

Fixed  and 
Round. 

Both  Ends 
Round. 

Timber 

I 

1.78 

4 

20000 

20  ooo 

20  ooo 

I 

1.78 

4 

30000 

30  ooo 

30000 

Wrought  Iron  

I 
40000 

1.78 

40  ooo 

4 
40000 

C»-,,1 

I 

1.78 

4 

24  ooo 

24  ooo 

24  ooo 

These  theoretical  values  of  q  are  smaller  than  the  empirical 
values  given  in  Art.  55,  except  that  for  steel.  Values  of/1 
computed  from  RITTER'S  formula  are  hence  generally  larger 
than  those  found  from  the  empirical  formulas  in  common  use. 
In  specifications,  however,  the  values  of  q  and  5  to  be  used 
are  generally  stated,  so  that  the  designer  is  free  from  the 
responsibility  of  selecting  them. 

It  should  be  noted  that  this  rational  formula,  having  been 
deduced  from  the  laws  which  govern  the  behavior  of  materials 
within  the  elastic  limit,  is  not  necessarily  true  for  cases  of 
rupture.  In  RANKINE'S  formula  (10)  the  unit-stress  5  may 
be  the  ultimate  strength  or  any  smaller  value,  but  in  RlTTER'S 
formula  5  cannot  exceed  the  elastic  limit  Se.  This  rational 
formula  cannot  hence  be  justified  or  disproved  because  it 
agrees  or  fails  to  agree  with  the  results  of  experiments  on  the 
rupture  of  columns. 

Prob.  1 02.  Compute  by  RITTER'S  formula  the  safe  steady 
load  for  a  hollow  cast-iron  column  with  fixed  ends,  the  out- 
side dimensions  being  4X5  inches,  the  inside  dimensions 
3X4  inches,  and  the  length  1 8  feet. 


132  COLUMNS   OR  STRUTS.  CH.  V. 

Prob.  103.  Find  by  RlTTER's  formula  the  size  of  a  square 
wooden  strut  with  fixed  ends,  and  12  feet  long,  to  carry  a 
load  of  100  ooo  pounds,  taking  .S  as  800  pounds  per  square 
inch. 


ART.  62.     ECCENTRIC  LOADS. 

In  all  that  precedes,  the  load  on  the  column  has  been  sup- 
posed to  be  applied  at  the  center  of  gravity  of  the  cross- 
section.  Let  now  the  case  be  considered  where  the  load  P 
is  applied  at  the  horizontal  distance  /  from  that  center  of 
gravity  and  in  the  same  vertical  plane  with  the  least  radius  of 
gyration  r.  The  reacting  load  at  the  other  end  is  similarly 
situated  with  respect  to  the  cross-section  at  that  end.  Let 
an  origin  be  taken  at  the  upper  end  of  the  column  and  let  x 
be  the  vertical  and  y  the  horizontal  co-ordinate  of  any  point 
in  the  elastic  curve.  The  column  having  round  ends,  the 
bending  moment  for  the  end  is  PP,  and  for  any  other  point 
*\p-\-y)-  Then  from  (5)  the  differential  equation  of  the 
elastic  curve  is 


where  the  negative  sign  is  used  because  the  curve  is  concave 
to  the  axis  of  x.     This  may  be  written, 

,       g  =-/?'(/  +  .,),     where     P  =  ^  ~, 

and  by  two  integrations  there  is  found, 
p   I      * 


as  may  be  proved  by  differentiating  the  last  equation  twice. 

The  deflection  at  the  middle  of  the  column  is  found  by 
making  x  =  £/  and  y  =  A,  whence 

"-  i). 


ART.  62.  ECCENTRIC   LOADS.  133 

From  this  equation  the  load  which  holds  the  deflected  column 
in  equilibrium  with  the  deflection  A  is  found  to  be 


which  reduces  to  EULER'S  formula  for  round-ended  columns 
when/=  o,  since  the  value  of  cos^o  is  \n. 

It  thus  appears  that  the  deflection  of  a  column  is  perfectly 
determinate  when  the  load  is  applied  eccentrically.  The 
above  formula  for  A  is,  however,  a  very  inconvenient  one  for 
practical  computations.  An  approximate  formula  giving 
closely  the  same  numerical  results  may  be  deduced  by  assum- 
ing that  the  curve  of  the  deflected  column  is  a  parabola;  the 
equation  of  this  curve  with  respect  to  an  origin  at  the  end  is 
y  =  (lx  —  x*)4.A/r.  Inserting  this  value  of  y  in  the  first  ex- 
pression of  this  article,  integrating  twice,  determining  the 
constants,  and  then  making  x  =  %l  and  y  =  A,  there  will  be 
found 

A  = 


9.6EI—PI" 

as  a  practical  formula  for  the  deflection  of  a  round-ended 
column  under  a  load  P  having  the  eccentricity/. 

Resuming  now  the  reasoning  of  Art.  55,  the  maximum 
compressive  unit-stress  S  on  the  concave  side  of  the  middle 
of  the  column  is 

P       Me       P,        pc  .    Ac 


and  from  this  S  may  be  computed  for  any  value  of  A.  By 
inserting  in  this  expression  the  above  value  of  A,  there 
results 


134  COLUMNS  OR  STRUTS.  CH.  V. 

pr 


PI    ,  pc    i  +  v  \ 
=  A(I  +  S'---&)> 


where   V'-= 


from  which  5  may  be  found  without  computing  A. 

For  example,  let  a  steel  column  with  round  ends  be  loaded 
with  168000  pounds.  Let  A  =  16.8  square  inches,  r  =  3.01 
inches,  £  =  4.45  inches,  /==  192  inches,  and  /=  I  inch. 
Then  P/A  =  10000  pounds  per  square  inch,  E  =  30  ooo  ooo 
pounds  per  square  inch,  /=  151.4  inches4.  Then  the  second 
formula  for  the  deflection  gives  A  =  0.197  inches,  and  the 
first  formula  for  5  gives  16900  pounds  per  square  inch.  In 
this  case  the  eccentric  load  increases  the  mean  unit-stress 
about  69  per  cent.  By  the  use  of  the  exact  formula  for  A  the 
same  numerical  values-  are  also  obtained. 

The  value  of  P/A  cannot  be  obtained  from  the  above 
formula  for  S,  since  it  is  contained  in  v.  By  transformation, 
however,  a  quadratic  equation  results  from  which  P/A  may 
be  computed  when  5  is  given.  A  more  convenient  way  will 
be  to  compute  5  for  an  assumed  value  of  P/A,  then  insert 
this  assumed  value  and  compute  again,  and  so  continue  until 
the  computed  and  assigned  values  of  5  agree,  which  they  will 
do  when  the  correct  value  of  P/A  has  been  found. 

Prob.  104.  Show  that  (/+  A)c  equals  r*  for  a  column  so 
deflected  that  there  is  no  stress  on  the  concave  side. 

Prob.  105.  A  wooden  strut  with  round  ends  is  18  feet 
long,  4  inches  square,  and  has  a  load  of  5000  pounds  applied 
half-way  between  the  center  and  corner  of  the  cross-section. 
What  is  the  direction  and  amount  of  deflection? 


ART.  63. 


THE  PHENOMENA  OF  TORSION. 


135 


Fig.  49- 


CHAPTER  VI. 
TORSION,  AND  SHAFTS   FOR  TRANSMITTING  POWER. 

ART.  63.    THE  PHENOMENA  OF  TORSION. 

Torsion  occurs  when  applied  forces  tend  to  cause  a  twisting 
of  a  body  around  an  axis.  Let  one  end  of  a  horizontal  shaft 
be  rigidly  fixed  and 
let  the  free  end  have 
a  lever  /  attached  at 
right  angles  to  its 
axis.  A  weight  P 
hung  at  the  end  of 
this  lever  will  twist 
the  shaft  so  that  fibers 
such  as  ab,  which  were 

originally  horizontal,  assume  a  spiral  form  ad  like  the  strands 
of  a  rope.  Radial  lines  such  as  cb  will  also  have  moved  through 
a  certain  angle  &cd. 

Experiments  have  proved,  that  if  P  be  not  so  large  as  to 
strain  the  material  beyond  its  elastic  limit,  the  angles  bed  and 
bad  are  proportional  to  P  and  that  on  the  removal  of  the  stress 
the  lines  cd  and  ad  return  to  their  original  positions  cb  and  ab. 
The  angle  bed  is  evidently  proportional  to  the  length  of  the 
shaft,  while  bad  is  independent  of  the  length.  If  the  elastic 
limit  be  exceeded  this  proportionality  does  not  hold,  and  if  the 
twisting  be  great  enough  the  shaft  will  be  ruptured.  These 
laws  are  but  a  particular  case  of  the  general  axioms  stated 
in  Art.  3. 

The  product  Pp  is  the  moment  of  the  force  P  with  respect  to 
the  axis  of  the  shaft,  /  being  the  perpendicular  distance  from 


136  TORSION  AND   SHAFTS.  CH.  VI. 

that  axis  to  the  line  of  direction  of  P,  and  is  called  the  twisting 
moment.  Whatever  be  the  number  of  forces  acting  at  the  end 
of  the  shaft,  their  resulting  twisting  moment  may  always  be 
represented  by  a  single  product  Pp. 

A  graphical  representation  of  the  phenomena  of  torsion  may 
be  made  as  in  Fig.  I,  the  angles  of  torsion  being  taken  as 
abscissas  and  the  twisting  moments  as  ordinates.  The  curve 
is  then  a  straight  line  from  the  origin  until  the  elastic  limit  of 
the  material  is  reached,  when  a  rapid  change  occurs  and  it 
soon  becomes  nearly  parallel  to  the  axis  of  abscissas.  The 
total  angle  of  torsion,  like  the  total  ultimate  elongation,  serves 
to  compare  the  relative  ductility  of  specimens. 

Prob.  1 06.  If  a  force  of  80  pounds  at  18  inches  from  the  axis 
twists  a  shaft  60°,  what  force  will  produce  the  same  result  when 
acting  at  4  feet  from  the  axis  ? 

Prob.  107.  A  shaft  2  feet  long  is  twisted  through  an  angle  of 
7  degrees  by  a  force  of  200  pounds  acting  at  a  distance  of  6 
inches  from  the  axis.  Through  what  angle  will  a  shaft  4  feet 
long  be  twisted  by  a  force  of  500  pounds  acting  at  a  distance 
of  1 8  inches  from  the  axis? 

ART.  64.    THE  FUNDAMENTAL  FORMULA  FOR  TORSION. 

The  stresses  which  occur  between  any  two  cross-sections  of 
a  bar  under  torsion  are  similar  to  those  of  shearing,  each  sec- 
tion tending  to  shear  off  from  the  one  ad- 
jacent to  it.  When  equilibrium  obtains  the 
external  twisting  moment  is  exactly  bal- 
anced by  the  sum  of  the  moments  of  these 
resisting  internal  stresses,  or, 

Resisting  moment  =  twisting  moment. 

The  law  governing  the  distribution  of  these 
5°.  internal  stresses  is  to  be  taken  the  same  as 

in  beams,  namely,  that  they  vary  directly  as  the  distance  from 


ART.  64.   THE   FUNDAMENTAL  FORMULA   FOR   TORSION.       137 

the  axis,  provided  that  the  elastic  limit  of  the  material  be  not 
exceeded. 

If  Pbe  the  force  acting  at  a  distance/  from  the  axis  about 
which  the  twisting  takes  place,  the  value  of  the  twisting  moment 
is  Pp.  To  find  the  resisting  moment,  let  c  be  the  distance  from 
the  axis  to  the  remotest  part  of  the  cross-section  where  the 
unit-shear  is  5,  .  Then  since  the  stresses  vary  as  their  distances 
from  the  axis, 

-—  =.  unit-stress  at  a  unit's  distance  from  axis, 

S  z 
-  =  unit-stress  at  a  distance  z  from  axis, 

/• 

—  —  =  total  stress  on  an  elementary  area  ay  ' 

—  —  =  moment  of  this  stress  with  respect  to  axis, 
—  a  ^  =  internal  resisting  moment. 


This  may  be  written  —  2az\     But  2  at?  is  the  polar  moment 


of  inertia  of  the  cross-section  with  respect  to  the  axis,  and  may 
be  denoted  by  /.     Therefore, 


(ii) 

which  is  the  fundamental  formula  for  torsion. 

The  analogy  of  formula  (11)  with  formula  (4)  for  the  flexure 
of  beams  will  be  noted.  Pp,  the  twisting  moment,  is  often  the 
resultant  of  several  forces,  and  might  have  been  expressed  by 
a  single  letter  like  the  M  in  (4).  By  means  of  (u)  a  shaft 
subjected  to  a  given  moment  may  be  investigated,  or  the 
proper  size  be  determined  for  a  shaft  to  resist  given  forces. 

Prob.  108.  Three  forces  of  120,  90,  and  70  pounds  act  at 
distances  of  6,  n,  and  8  inches  from  the  axis  and  at  different 


138  TORSION   AND    SHAFTS.  CH.  VI. 

distances  from  the  end  of  a  shaft,  the  direction  of  rotation  of 
the  second  force  being  opposite  to  that  of  the  others.  Find 
the  three  values  of  the  twisting  moment  Pp. 

Prob.  109.  A  circular  shaft  is  subjected  to  a  maximum 
shearing  unit-stress  of  2  ooo  pounds  when  twisted  by  a  force  of 
90  pounds  at  a  distance  of  27  inches  from  the  center.  What 
unit-stress  will  be  produced  in  the  same  shaft  by  two  forces  of 
40  pounds,  one  acting  at  21  and  the  other  at  36  inches  from 
the  center? 

ART.  65.     POLAR  MOMENTS  OF  INERTIA. 

The  polar  moment  of  inertia  for  simple  figures  is  readily 
found  by  the  help  of  the  calculus,  as  explained  in  works  on 
elementary  mechanics.  It  is  also  a  fundamental  principle 
that, 

/=/,  +  /,, 

where  /  is  the  polar  moment  of  inertia,  7,  the  least  and  72  the 
greatest  rectangular  moment  of  inertia  about  two  axes  passing 
through  the  center.  The  following  are  values  of  J  for  some 
of  the  most  common  cases. 

For  a  circle  with  a  diameter  d,  J  = , 

For  a  square  whose  side  is  d,  J  •=.  -^-, 

For  a  rectangle  with  sides  b  and  d,        /  ==  —  —  -J . 

12         12 

The  value  of  c  in  all  cases  is  the  distance  from  the  axis 
about  which  the  twisting  occurs,  usually  the  center  of  figure 
of  the  cross-section,  to  the  remotest  part  of  the  cross-section. 
Thus, 

For  a  circle  with  diameter  d,  c -=  %d, 

For  a  square  whose  side  is  d,  c  =  d  y^, 

For  a  rectangle  with  sides  b  and  d,     c  =  %  -\/&  -(-  d*. 


ART.  66.  THE  CONSTANTS  OF  TORSION.  139 

It  is  rare  in  practice  that  formulas  for  torsion  are  needed  for 
any  cross-sections  except  squares  and  circles. 

Prob.  no.  Find  the  values  of  J  and  c  for  an  equilateral  tri- 
angle whose  side  is  d. 

Prob.  in.  Find,  from  the  data  in  Art.  30,  the  values  of  J 
and  c  for  a  light  6  inch  /  section. 

ART.  66.    THE  CONSTANTS  OF  TORSION. 

The  constant  5S  computed  from  experiments  on  the  rupture 
of  shafts  by  means  of  formula  (u)  may  be  called  the  modulus 
of  torsion,  in  analogy  with   the  modulus  of  rupture  as  com 
puted  from  (4).     The  values  thus  found  agree  closely  with  the 
ultimate  shearing  unit-stress  given  in  Art.  7,  viz., 

For  timber,  S,  =    2  ooo  pounds  per  square  inch, 

For  cast  iron,          St  =  25  ooo  pounds  per  square  inch, 
For  wrought  iron,  5,  =  50  ooo  pounds  per  square  inch,  » 
For  steel,  S,  =  75  ooo  pounds  per  square  inch. 

By  the  use  of  these  average  values  it  is  hence  easy  to  compute 
from  (11)  the  load  P  acting  at  the  distance  /  which  will  cause 
the  rupture  of  a  given  shaft. 

The  coefficient  of  elasticity  for  shearing  may  be  computed 
from  experiments  on  torsion  in  the  following  manner.  Let  a 
circular  shaft  whose  length  is  /  and  diameter  d  be  twisted 
through  an  arc  6  by  the  twisting  moment  Pp.  Here  a  point 
on  the  circumference  of  one  end  is  twisted  relative  to  a  corre- 
sponding point  on  the  other  end  through  the  arc  0  or  through 
the  distance  \Qd,  so  that  the  detrusion  per  unit  of  length  is 

_ 


From  the  fundamental  definition  of  the  coefficient  of  elasticity 
Et  as  given  in  (2), 

_S,_2S,l 

*t?  7    ~8P 


140  TORSION  AND    SHAFTS.  CH.  VI. 

and  inserting  for  Ss  its  value  from  (u),  there  results, 

E  __  &Ppl 

'       ntid*  ' 

from  which  Et  can  be  computed  when  all  the  quantities  in  the 
second  member  have  been  determined  by  experiment,  pro- 
vided that  the  elastic  limit  of  the  material  be  not  exceeded. 
The  numerical  value  of  0  must  here  be  expressed  in  terms  of 
the  same  unit  as  n. 

Prob.  112.  What  force  P  acting  at  the  end  of  a  lever  24 
inches  long  will  twist  asunder  a  steel  shaft  1.4  inches  in 
diameter? 

Prob.  113.  An  iron  shaft  5  feet  long  and  2  inches  in  diam- 
eter is  twisted  through  an  angle  of  7  degrees  by  a  force  of 
5  ooo  pounds  acting  at  6  inches  from  the  center,  and  on  the  re- 
moval of  the  force  springs  back  to  its  original  position.  Find 
the  value  of  E  for  shearing. 

ART.  67.    SHAFTS  FOR  THE  TRANSMISSION  OF  POWER. 

Work  is  the  product  of  a  resistance  by  the  distance  through 
which  it  acts,  and  is  usually  measured  in  foot-pounds.  A  horse- 
power is  33000  foot-pounds  of  work  done  in  one  minute.  It 
is  required  to  determine  the  relation  between  the  horse-power 
H  transmitted  by  a  shaft  and  the  greatest  internal  shearing 
unit-stress  5,  produced  in  it. 

Let  a  shaft  making  n  revolutions  per  minute  transmit  H 
horse-power  .The  work  may  be  applied  by  a  belt  from  the  motor 
to  a  pulley  on  the  shaft,  then,  by  virtue  of  the  elasticity  and  re- 
sistance of  the  material  of  the  shaft,  it  is  carried  through  other 
pulleys  and  belts  to  the  working  machines.  In  doing  this  the 
shaft  is  strained  and  twisted,  and  evidently  S,  increases  with  H. 
Let  P  be  the  resistance  acting  at  the  circumference  of  the  pulley 
and/  the  radius  of  the  pulley.  In  making  one  revolution  the 


ART.  68.  ROUND  SHAFTS.  141 

force  Pacts  through  the  distance  2np  and  performs  the  work 
2npP,  and  in  n  revolutions  it  performs  the  work  27tpPn.  Then 
if  P  be  in  pounds  and  /  in  inches,  the  imparted  horse-power  is, 

H_       2npPn 
33  ooo  X  12 

The  twisting  moment  Pp  in  this  expression  may  be  expressed, 
as  in  formula  (11),  by  the  resisting  moment  -^-.  Hence  the 
equation  becomes, 

(12)  H  = 

198000^ 

This  is  the  formula  for  the  discussion  of  shafts  for  the  trans- 
mission of  power,  and  in  it  /  and  c  must  be  taken  in  inches  and 
5S  in  pounds  per  square  inch,  while  n  is  the  number  of  revolu- 
tions per  minute. 

Prob.  114.  A  wooden  shaft  6  inches  square  breaks  when 
making  40  revolutions  per  minute.  Find  the  horse-power  then 
probably  transmitted. 

ART.  68.    ROUND  SHAFTS. 

For  round  shafts  of  diameter  d,  the  values  of  J  and  c  are  to 
be  taken  from  Art.  65  and  inserted  in  the  last  equation,  giving, 

TT 

5,  =  321  ooo--—,         or 
na 

The  first  of  these  may  be  used  for  investigating  the  strengt- 
of  a  given  shaft  when  transmitting  a  certain  number  of  horse- 
power with  a  known  velocity.  The  computed  values  of  Ss, 
compared  with  the  ultimate  values  in  Art.  67,  will  indicate  the 
degree  of  security  of  the  shaft.  Here  d  must  be  taken  in 
inches  and  S,  will  be  in  pounds  per  square  inch. 

The  second  equation  may  be  used  for  determining  the  di- 
ameter of  a  shaft  to  transmit  a  given  horse-power  with  a  given 


142  TORSION   AND   SHAFTS.  CH.  VI. 

number  of  revolutions  per  minute.  Here  a  safe  allowable 
value  must  be  assumed  for  .Ss  in  pounds  per  square  inch,  and 
then  d  will  be  found  in  inches.  This  equation  shows  that  the 
diameter  of  a  shaft  varies  directly  as  the  cube  root  of  the 
transmitted  horse-power  and  inversely  as  the  cube  root  of  its 
velocity. 

Prob.  115.  Find  the  factor  of  safety  for  a  wrought  iron  shaft 
2^  inches  in  diameter  when  transmitting  25  horse-power  while 
making  100  revolutions  per  minute. 

Prob.  116.  Find  the  diameter  of  a  wrought  iron  shaft  to 
transmit  90  horse-power  with  a  factor  of  safety  of  8  when  mak- 
ing 250  revolutions  per  minute,  and  also  when  making  100 
revolutions  per  minute. 

ART.  69.    HOLLOW  SHAFTS. 

Hollow  forged  steel  shafts  are  now  coming  into  use  for  ocean 
steamers,  their  strength  being  greater  than  solid  shafts  of  the 
same  sectional  area.  If  D  be  the  exterior  and  d  the  interior 
diameter,  and  A  the  area  of  the  cross-section,  the  polar  moment 
of  inertia  is, 


and  the  discussion  of  any  case  can  be  made  by  formula  (12),  c 
being  replaced  by  %D. 

For  example,  let  it  be  required  to  determine  the  interior 
diameter  of  a  nickel-steel  shaft,  when  D  =  17  inches,  to  trans- 
mit 16  ooo  horse  power  at  50  revolutions  per  minute,  with  a 
stress  of  25  ooo  pounds  per  square  inch  on  the  exterior  circum- 
ference. Here  everything  is  given  except  d,  and  by  solution 
its  value  is  found  to  be  n  inches  nearly. 

Shafts  are  subject  to  flexural  stresses  due  to  their  own  weight 
and  to  applied  loads,  as  well  as  to  torsional  stresses.  The  effect 
of  these  will  be  discussed  in  Art.  76. 


ART.  70.  MISCELLANEOUS   EXERCISES.  H3 

Prob.  117.  Find  the  diameter  of  a  solid  shaft  for  the  con- 
ditions of  the  above  example,  and  compare  its  weight  with  that 
of  the  hollow  one. 

Prob.  1 1 8.  Find  the  horse-power  transmitted  by  a  hollow 
shaft,  when  D  —  15!  inches,  d  =  o>£  inches,  and  Ss=  12  500 
pounds  per  square  inch,  the  number  of  revolutions  per  minute 
being  50. 

ART.  70.    MISCELLANEOUS  EXERCISES. 

Exercise  8.  Make  experiments  to  verify  the  phenomena  of 
torsion  stated  in  Art.  63.  Show  by  your  experiments  that  the 
strength  of  a  round  shaft  varies  directly  as  the  cube  of  its 
diameter,  and  is  independent  of  its  length. 

Exercise  9.  Make  a  theoretical  investigation  to  ascertain  if 
the  strength  of  a  square  shaft  can  be  increased  by  cutting  off 
material  from  the  corners.  If  such  is  found  to  be  the  case 
write  an  essay  explaining  the  reasoning,  the  computations  and 
the  conclusion. 

Exercise  10.  Go  to  a  testing  room  and  inspect  THURSTON's 
testing  machine  for  torsion.  Ascertain  the  dimensions  and 
kind  of  specimens  tested  thereon.  Explain  with  sketches  the 
construction  of  the  machine,  the  method  of  its  use,  and  the 
torsion  diagrams.  State  how  the  quality  of  the  specimens  is 
inferred  from  the  torsion  diagrams. 

Prob.  119.  Compare  the  strength  of  a  square  shaft  with  that 
of  a  circular  shaft  of  equal  area. 

Prob.  1 20.  Compare  the  strengths  of  two  shafts  when 
stressed  to  their  elastic  limits;  the  first  shaft  is  solid,  21 
inches  in  diameter,  and  has  an  elastic  limit  of  25  ooo  pounds 
per  square  inch;  the  second  shaft  is  hollow,  18  inches  outside 
and  9  inches  inside  diameter,  and  its  elastic  limit  is  45  ooo 
pounds  per  square  inch. 


144  COMBINED   STRESSES.  CH.  VII. 


CHAPTER  VII. 
COMBINED  STRESSES. 

ART.  71.    COMBINED  TENSION  AND  COMPRESSION. 

Tensile  and  compressive  forces  acting  upon  a  bar  in  the 
direction  of  its  length,  produce  a  resultant  stress  equal  to  their 
numerical  difference,  which  may  be  either  tensile  or  compres- 
sive. This  case  is  of  frequent  occurrence  in  the  members  of 
bridge  trusses. 

A  tensile  force  acting  upon  a  bar  produces  a  tensile  unit- 
stress  5  and  unit-elongation  s.  It  is  found  by  experiment  that 
the  lateral  unit-contraction  of  the  bar,  when  5  is  within  the 
elastic  limit,  is  about  -J^,  and  hence  the  internal  compressive 
unit-stress  normal  to  the  length  of  the  bar  is  about  ^S.  Thus 
internal  stresses  may  exist  in  a  body  in  directions  which  do  not 
correspond  with  any  of  the  applied  exterior  forces.  In  general, 
if  s  be  any  unit-deformation,  the  corresponding  internal  unit- 
stress  is  sE  (Art.  4). 

If  three  tensile  forces  Pl ,  P^ ,  and  P3  act  normally  upon  the 
sides  of  a  rectangular  prism  whose  areas  are  Al ,  A^ ,  and  A3,  the 
unit-stresses  apparently  produced  upon  those  sides  are  Sl  = 
P1-^A1)S,  =  P^^-A,,  and  53  =  P3  -^  A, ;  but  the  real  effective 
internal  unit-stresses  are  much  smaller,  their  values,  by  the 
principle  of  the  last  paragraph,  being  Tl  =  Sl  —  J55  —  -J53 , 
T,  =  S,  -  to  -  iS, ,  and  Tt  =  S,  -  fS,  -  fSt.  These  for- 
mulas apply  when  some  or  all  of  the  external  forces  are  com- 
pressive as  well  as  tensile,  if  the  apparent  unit-stresses  be  taken 
positive  when  tension  and  negative  when  compression.  For 
example,  let  a  cube  whose  edge  is  unity  be  subject  to  a 


ART.  72.  STRESSES   DUE   TO   TEMPERATURE.  145 

compression  of  60  pounds  upon  two  opposite  sides  and  to  45 
pounds  tension  upon  two  other  opposite  sides,  the  third  pair 
of  sides  having  no  forces  applied  to  them.  Then  the  effective 
internal  unit-stress  normal  to  the  first  pair  of  sides  is  75  pounds 
compression,  that  for  the  second  pair  is  65  pounds  tension,  and 
that  for  the  third  pair  is  5  pounds  tension. 

Prob.  121.  A  common  brick  2X4X8  inches  is  subject  to 
compression  of  3  200  pounds  upon  its  top  and  bottom  faces,  500 
pounds  upon  its  sides,  and  60  pounds  upon  its  ends.  Find  the 
effective  internal  unit-stresses  in  the  three  directions. 

ART.  72.    STRESSES  DUE  TO  TEMPERATURE. 

If  a  bar  be  unstrained  it  expands  when  the  temperature  rises 
and  contracts  when  the  temperature  falls.  But  if  the  bar  be 
under  stress,  so  that  the  change  of  length  cannot  occur,  an  ad- 
ditional unit-stress  must  be  produced  which  will  be  equivalent 
to  the  unit- stress  that  would  cause  the  same  change  of  length 
in  the  unstrained  bar.  Thus  if  a  rise  of  temperature  elongates 
a  bar  of  length  unity  the  amount  s  when  free  from  stress,  it 
will  cause  the  unit-stress  S  =  sE  (see  Art.  4)  when  the  bar  is 
prevented  from  expanding  by  external  forces. 

Let  /  be  the  length  of  the  bar,  a  its  coefficient  of  linear  ex- 
pansion for  a  change  of  one  degree,  and  X  the  change  of  length 
due  to  the  rise  or  fall  of  /  degrees.  Then, 

;t  =  atl. 
and  the  unit-deformation  s  is, 

A 
s  =  -,  =  at. 

The  unit-stress  produced  by  the  change  in  temperature  hence  is, 

5  =  atE 

which  is  seen  to  be  independent  of  the  length  of  the  bar.  The 
total  stress  on  the  bar  is  then  AS. 


146  COMBINED   STRESSES.  CH.  VII. 

The  following  are  average  values  of  the  coefficients  of  linear 
expansion  for  a  change  in  temperature  of  one  degree  Fahren- 
heit. 

For  brick  and  stone,  a  =  o.ooo  oo  50, 

For  cast  iron,  a  =  o.ooo  oo  62, 

For  wrought  iron,  a  =  o.ooo  oo  67, 

For  steel,  a  =  o.ooo  oo  65. 

As  an  example  consider  a  wrought  iron  tie  rod  20  feet  in 
length  and  2  inches  in  diameter  which  is  screwed  up  to  a  ten- 
sion of  9  ooo  pounds  in  order  to  tie  together  two  walls  of  a 
building.  Let  it  be  required  to  find  the  stress  in  the  rod  when 
the  temperature  falls  10°  F.  Here, 

5  =  o.ooo  oo  67  X  10  X  25  ooo  ooo  =  i  675  pounds. 
The  total  tension  in  the  rod  now  is, 

9  ooo  +  3.14  X  i  675  =  14  ooo  pounds. 
Should  the  temperature  rise  10°  the  tension  in  the  rod  would 

become, 

9  ooo  —  3.14  X  i  675  =  4000  pounds. 

In  all  cases  the  stresses  caused  by  temperature  are  added  or 
subtracted  to  the  tensile  or  compressive  stresses  already 
existing. 

Prob.  122.  A  cast  iron  bar  is  confined  between  two  immovable 
walls.  What  unit-stress  will  be  produced  by  a  rise  of  40°  in 
temperature  ? 

ART.  73.    COMBINED  TENSION  AND  FLEXURE. 

Consider  a  beam  in  which  the  flexure  produces  a  unit-stress 
Sl  at  the  fiber  on  the  tensile  side  most  remote  from  the  neutral 
axis.  Let  a  tensile  stress  P  be  then  applied  to  the  ends  of  the 

bar  uniformly  distributed  over  the  cross-section  A.     The  ten- 

p 
sile  unit-stress  at  the  neutral  surface  is  then    -  and  all  the 

longitudinal  stresses  due  to  the  flexure  are  increased  by  this 


ART.  73.  COMBINED   TENSION   AND    FLEXURE.  147 

P 

amount.     Then  5  =  —  +  5,  is  an  approximate  value  of  the 

A 

maximum  tensile  unit-stress. 

In  designing  a  beam  under  combined  tension  and  flexure  the 

p 

dimensions  must  be  so  chosen  that  —  7  -j-  5,  shall  not  exceed 

si 

the  proper  allowable  working  unit-stress.  For  instance,  let  it 
be  required  to  find  the  size  of  a  square  wooden  beam  of  12 
feet  span  to  hold  a  load  of  300  pounds  at  the  middle  while 
under  a  longitudinal  stress  of  2000  pounds,  so  that  the  maxi- 
mum tensile  unit-stress  may  be  about  I  ooo  pounds  per  square 
inch.  Let  d  be  the  side  of  the  square.  From  formula  (4), 

c  _  6J/_  6  X  150  X  72 
'"          ~- 


Then  from  the  conditions  of  the  problem, 
2  ooo  ,  64  800  _ 

~d^   ~^~      ^ 

from  which  results  the  cubic  equation, 

d*  -  2^=64.8, 
whose  solution  gives  for  d  the  value  4.25  inches. 

In  investigating  a  beam  under  combined  tension  and  flexure 
the  maximum  combined  unit-stress  5  is  computed,  and  the 
factor  of  safety  found  by  comparing  it  with  the  ultimate 
tensile  strength  of  the  material.  The  method  here  given  is 
approximate;  more  accurate  methods  are  in  Art.  118. 

Prob.  123.  A  heavy  12-inch  I  beam  of  6  feet  span  carries  a 
uniform  load  of  200  pounds  per  linear  foot,  besides  its  own 
weight,  and  is  subjected  to  a  longitudinal  tension  of  80000 
pounds.  Find  the  factor  of  safety  of  the  beam. 

Prob.  124.  What  I  beam  of  12  feet  span  is  required  to  carry 
a  uniform  load  of  200  pounds  per  linear  foot  when  subjected 
to  a  tension  of  50000  pounds,  the  maximum  tensile  stress  at 
the  dangerous  section  to  be  loooo  pounds  per  square  inch  ? 


148  COMBINED   STRESSES.  CH.  VII. 

ART.  74.    COMBINED  COMPRESSION  AND  FLEXURE. 

Consider  a  beam  in  which  the  flexure  produces  a  unit-stress 
.S,  in  the  fiber  on  the  compressive  side  most  remote  from  the 
neutral  axis.  Let  a  compressive  stress  P  be  applied  in  the  direc- 
tion of  its  length  uniformly  over  the  cross-section  A.  Then  at 

p 
the  neutral  surface  the  unit-stress  is    —  and  at  the  remotest 

A 
p 

fiber  it  is  -—  +  51,.    The  discussion  of  this  case  is  hence  exactly 
A 

similar  to  that  of  the  last  article.  If  the  beam  is  short  the 
total  working  unit-stress  is  to  be  taken  as  for  a  short  prism  ; 
if  long  it  should  be  derived  from  RANKINE'S  formula  for 
columns. 

The  method  of  investigation  explained  in  this  and  the  pre- 
ceding article  is  the  one  ordinarily  used  in  practice  on  account 
of  the  complexity  of  the  formulas  which  result  from  the  strict 
mathematical  determination  of  the  moments  of  the  applied 
forces.  Although  not  exact  the  method  closely  approximates 
to  the  truth,  giving  values  of  the  stresses  a  little  too  large  for 
the  case  of  tension  and  a  little  too  small  for  the  case  of 
compression.  (See  Arts.  117  and  118.) 

A  rafter  of  a  roof  is  a  case  of  combined  compression  and 

flexure.  Let  b  be  its  width, 
d  its  depth,  /  the  length, 
w  the  load  per  linear  unit, 
and  0  the  angle  of  inclina- 
tion. To  find  the  horizon- 
tal reaction  H  the  center 
of  moments  is  to  be  taken 
Flg"  5I<  at  the  lower  end,  and 

TT       ,     .         ,  j     I  COS  0,  rr          Wl 

H  ,  I  sin  0  =  wl . — ,     whence     ff=  —  cot  0. 

2  -  2 


ART.  74.     COMBINED    COMPRESSION   AND    FLEXURE.  149 

For  any  section  whose  distance  from  the  upper  end  is  x,  the 
flexural  unit-stress  now  is  from  (4), 

c  _  6M  _  6(Hx  sin  0  —  \wx*  cos  0) 

^~W  ~~bd^~ 

and  the  uniform  compressive  unit-stress  is, 
c  _  H  cos  0  -f-  wx  sin  0 

6>-        ~w"~ 

The  total  compressive  unit-stress  on  the  upper  fiber  hence  is, 
c        o         o       3W/COS0,,         ^   ,   w/cot0cos0 
=  5'  +  5'~   -~^~(lX  ~- 


bd 
This  can  be  shown  to  be  a  maximum  when 

*  =  i/+f</  tan  0, 
and  substituting  this,  the  maximum  unit-stress  is, 

«  _  3  wl*  cos  0      w/cosec  0      wsin  0  tan  0 
a  "  ~"~ 


which  formula  may  be  used  to  investigate  or  to  design  rafters 
subject  to  uniform  loads. 

In  any  inclined  rafter  let  P  denote  all  the  load  above  a  sec- 
tion distant  x  from  the  upper  end.     Then  reasoning  as  before 
the  greatest  unit-stress  for  that  section  is  found  to  be, 
c  _  Me  .PsmQ.H  cos  0 
:7"     ~A~         ~A~> 
from  which  Sx  may  be  computed  for  any  given  case. 

Prob.  125.  A  roof  with  two  equal  rafters  is  40  feet  in  span 
and  1  5  feet  in  height.  The  wooden  rafters  are  4  inches  wide 
and  each  carries  a  load  of  450  pounds  at  the  center.  Find 
the  depth  of  the  rafter  so  that  5  may  be  700  pounds  per 
square  inch. 

Prob.  126.  A  wooden  beam  10  inches  wide  and  8  feet  long 
carries  a  uniform  load  of  500  pounds  per  linear  foot  and  is  sub- 
jected to  a  longitudinal  compression  of  40  ooo  pounds.  Find 
the  depth  of  the  beam  so  that  the  maximum  working  unit-stress 
may  be  about  800  pounds  per  square  inch. 


ISO 


COMBINED   STRESSES. 


CH.  VIL 


ART.  75    SHEAR  COMBINED  WITH  TENSION  OR  COMPRESSION. 

Let  a  bar  whose  cross-section  is  A  be  subjected  to  the  longi- 
tudinal tension  or  compression  P  and  at  the  same  time  to  a 
shear  Fat  right  angles  to  its  length.  The  longitudinal  unit- 

p 
stress  is  -—  which  may  be  denoted  by/,  and  the  shearing  unit- 

A 

stress  is  -  which  may  be  denoted  by  v.     It  is  required  to  find 

the  maximum  unit-stresses  produced  by  the  combination  of  p 
and  v.  In  the  following  demonstration  P  will  be  regarded  as 
a  tensile  force,  although  the  reasoning  and  conclusions  apply 
equally  well  when  it  is  compressive. 

Consider  an  elementary  cubic  particle  with  edges  one  unit 
in  length  acted,  upon  by  the  horizontal  tensile  force  p  and  p., 
and   by   the   vertical    shear   v  and  v,  as 
shown  in  Fig.  52.     These  forces  are  not  in 
P  y       equilibrium  unless  a  horizontal  couple  be 

>•   applied  as  in  the  figure,  each  of  whose 

forces  is  equal  to  v.     Therefore  at  every 
Fig.  52-  point  of  a  body  under  vertical  shear  there 

exists  a  horizontal  shear,  and  the  horizontal  shearing  unit- 
stress  is  equal  to  the  vertical  shearing  unit-stress. 

Let  a  parallelopipedal  element  have  the  length  dm,  the 
height  dn,  and  a  width  of  unity.  The  tensile  iorc.Qp.dn  tends 
to  pull  it  apart  longitudinally. 
The  vertical  shear  vdn  tends  to 
cause  rotation  and  this  is  resisted, 
as  shown  above,  by  the  horizon- 
tal shear  vdm.  These  forces 
maybe  resolved  into  rectangular 
components  parallel  and  perpen- 
dicular to  the  diagonal  dz,  as  shown  in  Fig.  53.  The  compo- 
nents parallel  to  the  diagonal  form  a  shearing  force  sdz,  and 


\ 


v.dn 


dm 


v.dn 


dn 


\ 


t.dz 


Fig.  53. 


ART.  75.  SHEAR   COMBINED   WITH   TENSION.  I$I 

those  perpendicular  to  it  a  tensile  force  tdz,  s  being  the  shear- 

ing and  t  the  tensile  unit-stresses.  Let  0  be  the  angle  between 

dz  and  dm.     The  problem  is  first  to  state  expressions  for  sdz 
and  tdz  in  terms  of  0,  and  then  to  determine  the  value  of  0, 

or  the  ratio  of  dm  to  dn,  which  gives  the  maximum  values 
of  s  and  t. 

By  simple  resolution  of  forces, 

sdz  =  pdn  cos  0  -[~  vdm  cos  0  —  vdn  sin  0, 
/*/£  =  ^/#  sin  0  -|-  w/w  sin  0  -\-  vdn  cos  0. 

Divide  each  of  these  by  dz,  for  -=-  put  its  value  sin  0  and  for 

-T-  its  value  cos  0.     Then  the  equations  take  the  form, 

s  =  p  sin  0  cos  0  -|-  z/(cosa  0  —  sina  0), 
t  =.  p  sin2  0  -|-  2z»  sin  0  cos  0. 

1'hese  may  be  written, 

s  =  %p  sin  20  -f-  z;  cos  20, 

/  =  £/(i  —  cos  20)  +  ?'  sin  20. 


By  placing  the  first  derivative  of  each  of  these  equal  to  zero  it 
is  found  that, 

P 

s  is  a  maximum  when  tan  20  =  —  , 

2it 

2V 

t  is  a  maximum  when  tan  20  =  --  . 

/ 

Expressing  sin  20  and  cos  20  in  terms  of  tan  20  and  inserting 
them  in  the  above  the  following  values  result  : 


jm«.,==±^+ 

/  max.  t  =  irf  +  */„* 


These  formulas  apply  to  the  discussion  of  the  internal  stresses 
in  beams,  as  well  as  to  combined  longitudinal  stress  and  vertical 
shear  directly  applied  by  external  forces.  If  p  is  tension  /  is 


152  COMBINED    STRESSES.  CH.  VII. 

tension,  if/  is  compression  /  is  also  compression.  If  when/  is 
tension  the  negative  sign  be  used  before  the  radical,  the  re- 
sultant value  of  t  is  the  maximum  compressive  unit-stress. 

Prob.  127.  A  bolt  f-inch  in  diameter  is  subjected  to  a  tension 
of  2  ooo  pounds  and  at  the  same  time  to  a  cross  shear  of  3  ooo 
pounds.  Find  the  maximum  tensile  and  shearing  unit-stresses 
and  the  directions  they  make  with  the  axis  of  the  bolt. 


ART.  76.    COMBINED  FLEXURE  AND  TORSION. 

This  case  occurs  when  a  shaft  for  the  transmission  of  power 
is  loaded  with  weights.  Let  5  be  the  greatest  flexural  unit- 
stress  computed  from  (4)  and  Ss  the  torsional  shearing  unit- 
stress  computed  from  (12)  or  by  the  special  equations  of  Arts. 
67  and  68.  Then,  according  to  the  last  article,  the  resultant 
maximum  unit-stresses  are, 


max.  ten.  or  comp.  /  —  -J5+  y  5/ 
max.  shear  s  =  ± 


For  wrought  iron  or  steel  it  is  usually  necessary  to  regard  only 
the  first  of  these  unit-stresses,  but  for  timber  the  second  should 
also  be  kept  in  view. 

For  example,  let  it  be  required  to  find  the  factor  of  safety  of 
a  wrought  iron  shaft  3  inches  in  diameter  and  12  feet  between 
bearings,  which  transmits  40  horse-power  while  making  120 
revolutions  per  minute,  and  upon  which  a  load  of  800  pounds 
is  brought  by  a  belt  and  pulley  at  the  middle.  Taking  the 
shaft  as  fixed  over  the  bearings  the  flexural  unit-stress  is, 


5=  — -js  =  5  40°  pounds  per  square  inch. 
Ttd 

From  Art.  68  the  torsional  unit-stress  is, 

IT 

Ss  =  321  ooo  --j-3  =  4  ooo  pounds  per  square  inch. 


ART.  /6.  COMBINED   FLEXURE  AND   TORSION.  153 

The  maximum  tensile  and  compressive  unit-stress  now  is, 

/  =  2  700  4-  ,y/4  ooo2  +~2~70oa  =  7  600  pounds  per  square  in. 
and  the  factor  of  safety  is  hence  over  7. 

As  a  second  example,  let  it  be  required  to  find  the  size  of 
a  square  wooden  shaft  for  a  water-wheel  weighing  3  ooo  pounds 
which  transmits  8  horse-power  while  making  20  revolutions  per 
minute.  The  length  of  the  shaft  is  16  feet,  and  one-third  of 
the  weight  is  concentrated  at  the  center  and  the  remainder  is 
equally  divided  between  two  points,  each  6  feet  from  the  center. 
Here  the  greatest  flexural  unit-stress  is, 

_  6(1  500  X  96  —  i  ooo  X  72)  _  432000 
~~d*~  d*     ' 

and  from  Art.  67  the  torsional  unit-stress  is, 
_  267  500  X  8  _  107000 

•J,  7". 7T . 

2od  d 

From  the  formula  of  the  last  Article  the  combined  tensile  or 
compressive  unit-stress  is, 

470400 
d3 

Now  if  the  working  value  of  /  be  taken  at  600  pounds  per 
square  inch  the  value  of  d  will  be  about  9  inches.  From  for- 
mula (13)  also 

s=  254400 

and  if  the  working  value  of  s  be  taken  at  1 50,  the  value  of  d,  is 
found  to  be  about  12  inches.  The  latter  value  should  hence 
be  chosen  for  the  size  of  the  shaft. 

,.    By  similar  reasoning  it  may  be  proved  that  the  formula  for 
finding  the  diameter  of  a  round  iron  shaft  is, 

\6M      16     I M*       402  500  ooo//'8 


154  COMBINED   STRESSES.  CH.  VIL 

where  M  is  the  maximum  bending  moment  of  the  transverse 
forces  in  pound-inches,  H  the  number  of  transmitted  horse- 
power, n  the  number  of  revolutions  per  minute,  and  /  the 
safe  allowable  tensile  or  compressive  working  strength  of  the 
material. 

Prob.  128.  Find  the  factor  of  safety  for  the  data  of  Prob. 
115  when  the  shaft  is  in  bearings  12  feet  apart  and  carries  a 
load  of  200  pounds  at  the  middle. 

ART.  77.    COMBINED  COMPRESSION  AND  TORSION. 

In  the  case  of  a  vertical  shaft  the  torsional  unit-stress  5,  com- 
bines with  the  direct  compressive  stress  due  to  the  weights 
upon  the  shaft,  and  produces  a  resultant  compression  t  and 
shear  s.  From  formulas  (13)  the  combined  unit-stresses  are, 


s  ~  v  S*  +  i-S;1. 

The  use  of  these  is  the  same  as  those  of  the  last  Article,  S9 
being  found  from  the  formulas  of  Chapter  VI,  while  Sc  is  com- 
puted from  formula  (i)  if  the  length  of  the  shaft  be  less  than 
ten  times  its  diameter  and  from  (10)  for  greater  lengths. 

In  order  to  prevent  vibration  and  flexure  it  is  usual  to  place 
bearings  at  frequent  intervals  on  a  vertical  shaft  so  that  prob- 
ably the  use  of  formula  (10)  will  rarely  be  required,  particularly 
if  /  be  taken  at  a  low  value.  For  a  round  shaft  the  expression 
for  t  becomes, 

/  =  — -'        7~        ~^' 


in  which  P  is  the  load.     From  this  the  diameter  d  may  be  found 
when  /  and  the  other  data  are  given. 

Prob.   129.  A  vertical  shaft,  weighing  with   its  loads   6000 


ART.  78.      HORIZONTAL  SHEAR  IN  BEAMS.  1 55 

pounds,  is  subjected  to  a  twisting  moment  by  a  force  of  300 
pounds  acting  at  a  distance  of  4  feet  from  its  center.  If  the 
shaft  is  wrought  iron,  4  feet  long  and  2  inches  in  diameter,  find 
its  factor  of  safety. 

Prob.  130.  Find  the  diameter  of  a  short  vertical  steel  shaft 
to  carry  loads  amounting  to  6  ooo  pounds  when  twisted  by  a 
force  of  300  pounds  acting  at  a  distance  of  4  feet  from  the 
center,  taking  the  unit-stress  against  compression  as  10000  and 
against  shearing  as  7  ooo  pounds  per  square  inch. 

ART.  78.    HORIZONTAL  SHEAR  IN  BEAMS. 

The  common  theory  of  flexure  as  presented  in  Chapters  III 
and  IV  considers  that  the  internal  stresses  at  any  section  are 
resolved  into  their  horizontal  and  vertical  components,  the 
former  producing  longitudinal  tension  and  compression  and 
the  latter  a  transverse  shear,  and  that  these  act  independently 
of  each  other.  Formula  (3)  supposes  further  that  the  vertical 
shear  is  uniformly  distributed  over  the  cross-section  of  the 
beam.  A  closer  analysis  will  show  that  a  horizontal  shear 
exists  also  and  that  this,  together  with  the  vertical  shear, 
varies  in  intensity  from  the  neutral  surface  to  the  upper  and 
lower  sides  of  the  beam.  It  is  well  known  that  a  pile  of  boards 
which  acts  like  a  beam  deflects  more  than  a  solid  timber  of  the 
same  depth,  and  this  is  largely  due  to  the  lack  of  horizontal 
resistance  between  the  layers.  The  common  theory  of  flexure 
in  neglecting  the  horizontal  shear  generally  errs  on  the  side  of 
safety.  In  a  few  experiments  however  beams  have  been  known 
to  crack  along  the  neutral  surface  and  it  is  hence  desirable  to 
investigate  the  effect  of  horizontal  shear  in  tending  to  cause 
rupture  of  that  kind.  That  a  horizontal  shear  exists  simulta- 
neously with  the  vertical  shear  is  evident  from  the  considera- 
tions in  Art.  75. 

Let  Fig.  54  represent  a  portion  of  a  bent  beam  of  uniform 


156 


COMBINED   STRESSES. 


CH.  VII. 


-r  "      v  —  y 

*%    ^' 

t 
\ 

m 

m' 

—  -*.  ^ 

1 

i 

•      fr- 

•—  —  ~-  

section.  Let  a  rectangular  notch  nmpq  be  imagined  to  be  cut 
into  it,  and  let  forces  be  applied  to  it  to  preserve  the  equilib- 
rium. Let  H  be  the  sum  of  all  the  horizontal  components  of 

these  forces  act- 
ing on  mn  and  H' 
the  sum  of  those 
acting  on  qp. 
Now  H'  is  greater 
or  less  than  H, 
Fig'54>  hence  the  differ- 

ence H'  —  H  must  act  along  mq  as  a  horizontal  shear.  Let  the 
distance  mq  be  dx,  the  thickness  mm!  be  £,  and  the  area  mqmm' 
be  at  a  distance  c'  above  the  neutral  surface.  Let  c  be  the  dis- 
tance from  that  neutral  surface  to  the  remotest  fiber  where  the 
unit-stress  is  5.  Let  a  be  the  cross-section  of  any  fiber.  Let 
M  be  the  bending  moment  at  the  section  mn  and  M'  that  at 
the  section  qp.  Now  from  the  fundamental  laws  of  flexure, 

-  =  unit-stress  at  a  unit's  distance  from  neutral  surface, 
c 

c- 

-  y  —  unit-stress  at  distance  y  from  neutral  surface, 


aSy  _ 
c 


=  total  stress  on  fiber  a  at  distance  y, 


=  sum  °f  horizontal  stresses  between  m  and  «. 


M 


The  value  of  H  hence  is,  since  -  —  --  , 


and  likewise  for  the  other  section, 


ART.  78.  HORIZONTAL   SHEAR   IN   BEAMS.  157 

The  horizontal  shear  therefore  is  expressed  by 

Mr  -M 
H'  -H=-  —  -  ^ay. 

Now  since  the  distance  mq  is  dx,  the  value  of  M*  —  M  is  dM. 
Also  if  Sk  be  the  horizontal  shearing  unit-stress  upon  the  area 
bdx  the  value  of  H1  —  H  is  SJbdx.  Hence, 

dM 


Again  from  Art.  45  it  is  plain  that  —  -  -  is  the  vertical  shear  V 
at  the  section  under  consideration.     Therefore, 

(14)  Sh  =  --^ayi 

lo 

is  the  formula  for  the  horizontal  shearing  unit-stress  at  any 
point  of  any  section  of  the  beam. 

This  expression  shows  that  the  horizontal  shearing  unit-stress 
is  greatest  at  the  supports,  and  zero  at  the  dangerous  section 
where  V  is  zero.  The  summation  expression  is  the  statical 
moment  of  the  area  mm'nn'  with  reference  to  the  neutral  axis  ; 
it  is  zero  when  cf  =  c,  and  a  maximum  when  c'  =  o.  Hence 
the  longitudinal  unit-shear  is  zero  at  the  upper  and  lower  sides 
of  the  beam  and  is  a  maximum  at  the  neutral  surface.  The 
formula  for  the  maximum  horizontal  shearing  unit-stress  at  any 
section  therefore  is, 


Here  /  is  the  moment  of  inertia  of  the  whole  cross-section  with 
reference  to  the  neutral  axis  (Art.  23),  b  is  the  width  of  the 
beam  along  the  neutral  surface,  and  ^C0ay\s  the  statical  mo- 
ment of  the  area  of  the  part  of  the  cross-section  on  one  side  of 
the  neutral  axis.  Let  Al  be  the  area  of  the  cross-section  on 


158  COMBINED    STRESSES.  CH.  VII. 

one  side  of  the  neutral  axis  and  cl  the  distance  of  its  center  of 
gravity  from  that  axis;  then  2c0ay  =  A^,  and  the  formula 

becomes, 

VA  c         A  ,  - 
(I4y  S.=   -j±±', 

k      i        u*rvWH\  A£^Mq   O 

which  gives  the  maximum  shearing  unit-stress,  both  horizontal 
and  vertical,  at  the  neutral  surface.  The  mean  unit-stress  given 
by  (3)  is  always  less  than  this  maximum. 

For  a  rectangular  beam  of  breadth  b  and  depth  d,  the  valu* 

.   bd*  bd    d      bd*      _, 

of  /  is ,  and  A.c.  =  —  .  —  = .     Then, 

12  248 


By  inserting  in  this  the  value  of  V  for  any  section  the  co*re- 
sponding  value  of  5,  at  the  neutral  surface  is  found.  In  this 
particular  instance  it  is  seen  that  the  approximate  formula  (3) 
gives  values  of  5S  which  are  33  per  cent  lower  than  the  true 
maximum  value. 

Prob.  131.  In  the  Journal  of  the  Franklin  Institute  for  Feb- 
ruary, 1883,  is  detailed  an  experiment  on  a  spruce  joist  3j  X  12 
inches  and  14  feet  long,  which  broke  by  tension  at  the  middle 
and  afterwards  by  shearing  along  the  neutral  axis  at  the  end 
when  loaded  at  the  middle  with  12  545  pounds.  Find  the  ten- 
sile and  shearing  unit-stresses. 

• 
ART.  79.    MAXIMUM  INTERNAL  STRESSES  IN  BEAMS. 

From  the  last  Article  it  is  evident  that  at  every  point  of  a 
beam  there  exists  a  horizontal  unit-shear  of  the  intensity  Sh  and 
also  a  vertical  unit-shear  of  the  same  intensity,  whose  value  is 
given  by  (14).  At  every  point  there  also  exists  a  longitudinal 
tension  or  compression  which  may  be  computed  from  (4)  with 
the  aid  of  the  principle  that  these  stresses  vary  directly  as  their 


ART.  79.     MAXIMUM   INTERNAL   STRESSES   IN   BEAMS.  159 

distances  from  the  neutral  axis.  Let  v  denote  the  unit-shear 
thus  determined  and  /  the  tensile  or  compressive  unit-stress. 
Then  from  Art.  75  the  maximum  unit-shear  at  that  point  is, 


and  it  makes  an  angle  0  with  the  neutral  surface  such  that, 

tan  2<t>=^_. 

2V 

Also  the  maximum  tensile  or  compressive  unit-stress  at  that 
point  is, 


and  it  makes  an  angle  0  with  the  neutral  surface  such  that, 

tan  20=-^. 
P 

From  these  formulas  the  lines  of  direction  of  the  maximum 
stresses  may  be  traced  throughout  the  beam. 

For  the  maximum  shear  v  is  greatest  and  p  is  zero  at  the 
neutral  surface,  while  v  is  zero  and  p  is  greatest  at  the  upper 
and  lower  surfaces.  Hence  for  the  neutral  surface  0  is  o,  it 
increases  with  p,  and  becomes  45°  at  the  upper  and  lower 

surfaces. 

For  the  maximum  tension  t  is  greatest  and  equal  to  p  on 
the  convex  side  where  v  =  o  and  6  =  o.  As  the  neutral  sur- 
face is  approached  v  increases,  /  decreases,  and  0  increases. 
At  the  neutral  surface  v  is  greatest,  /  is  zero,  and  0  =  —  45°. 
Here  the  maximum  tension  and  compression  are  each  equal 
to  v. 

For  the  maximum  compression  in  like  manner  6  is  o°  at  the 
concave  surface  and  45°  at  the  neutral  surface.  The  lines  of 
maximum  tension  if  produced  beyond  the  neutral  surface  would 
evidently  cut  those  of  maximum  compression  at  right  angles 
and  be  vertical  at  the  concave  surface. 


l6o  COMBINED   STRESSES.  CH.  VII. 

The  following  figure  is  an  attempt  to  represent  the  lines 
which  indicate  the  directions  of  the  maximum  unit-stress  in  a 

beam.  The  full  lines 
above  the  neutral 
surface  are  those  of 
maximum  compres- 
sion, while  those  be- 
low are  maximum 
tension.  The  broken 
lines  are  those  of 
Flg'55'  maximum  shear.  On 

any  line  the  intensity  of  stress  varies  with  the  inclination,  being 
greatest  where  the  line  is  horizontal  and  least  where  its  inclina- 
tion is  45°.  The  lines  of  maximum  shear  cut  those  of  maxi- 
mum tension  and  compression  at  angles  of  45°.  The  lines  of 
maximum  tension  above  the  neutral  surface  and  those  of  maxi- 
mum compression  below  it  are  not  shown  ;  if  drawn  they  would 
cut  the  others  at  right  angles  and  become  vertical  at  the  upper 
and  lower  edges  of  the  beam. 

It  appears  from  the  investigation  that  the  common  theory  of 
flexure  gives  the  horizontal  unit-stress  correctly  at  the  dan- 
gerous section  of  a  simple  beam  where  the  vertical  shear  is 
zero.  At  other  sections  the  stress  S  as  computed  from  (4)  is 
correct  for  the  remotest  fiber,  but  for  other  fibers  the  unit-stress 
t  is  greater.  It  is  hence  seen  that  the  main  practical  value  of 
the  theory  of  internal  stress  is  in  showing  that  the  intensity  of 
the  shear  varies  throughout  the  cross-section  of  the  beam.  For 
a  restrained  beam,  where  the  vertical  shear  suddenly  changes 
sign  at  the  dangerous  section,  the  common  theory  gives  the 
horizontal  stress  5  correctly  for  the  remotest  fiber  only,  and 
it  might  be  possible  in  some  forms  of  cross-sections  for  the 
maximum  stress  /  to  be  slightly  greater  than  S  for  a  fiber 
nearer  to  the  neutral  surface.  All  that  has  here  been  deduced 


ART.  79.     MAXIMUM   INTERNAL  STRESSES   IN   BEAMS.  l6l 

justifies  the  validity  of  the  common  theory  of  flexure  as  a 
correct  guide  in  the  practical  design  and  investigation  of 
beams. 

Prob.  132.  A  joist  fixed  at  both  ends  is  3  X  12  inches  and 
12  feet  long,  and  is  strained  by  a  load  at  the  middle,  so  that 
the  value  of  5  as  computed  from  (4)  is  4000  pounds  per  square 
inch.  Find  the  value  of  t  for  points  over  the  support  distant 
3,  4,  and  5  inches  from  the  neutral  surface. 

Prob.  133.  Show,  fora  point  between  the  neutral  surface  and 
the  convex  side,  that  there  exists  a  maximum  compression  as 
well  as  a  maximum  tension.  Deduce  an  expression  for  the 
value  of  this  maximum  compression  and  its  direction.  Draw 
a  figure  showing  the  curves  over  the  entire  beam  for  both  these 
stresses. 


1 62  THE   STRENGTH   OF   MATERIALS.  CH.  VIII, 


CHAPTER    VIII. 
THE    STRENGTH     OF     MATERIALS. 

ART.  80.     MEAN  CONSTANTS. 

The  term  '  strength  of  materials  '  is  generally  understood 
to  refer  to  stresses  caused  by  slowly  applied  loads,  where  the 
conditions  are  those  of  statics  only.  The  ultimate  tensile 
strength  of  timber,  for  example,  is  obtained  by  placing  a  speci- 
men, like  that  shown  in  Fig.  2  of  Art.  8,  in  a  testing  machine 
and  pulling  it  by  a  force  which  gradually  increases  until  rup- 
ture occurs.  When  forces  are  applied  suddenly  or  with  shock, 
the  condition  is  one  of  work  or  resilience  (see  Chapter  IX). 

The  following  tables  recapitulate  the  mean  values  of  the  con- 
stants of  the  strength  of  materials  which  have  been  given  in 
the  preceding  pages.  It  is  here  again  repeated  that  these 
values  are  subject  to  wide  variations  dependent  on  the  kind 
and  quality  of  the  material,  and  for  many  other  reasons.  Tim- 
ber, for  instance,  varies  in  strength  according  to  the  climate 
where  grown,  the  soil,  the  age  of  the  tree,  the  season  of  the 
year  when  cut,  the  method  and  duration  of  the  process  of 
seasoning,  the  part  of  the  tree  used,  the  knots  and  wind  shakes, 
the  form  and  size  of  the  test  specimen,  and  the  direction  of  its 
fibers,  so  that  it  is  a  difficult  matter  to  state  definite  numerical 
values  concerning  its  elasticity  and  strength.  The  quality  of 
the  material  causes  a  yet  wider  variation,  so  wide  in  fact  that 
in  some  cases  testing  machines  alone  could  scarcely  distinguish 
between  wrought  iron  and  steel ;  for  while  the  higher  grades  of 
steel  have  much  greater  strength  than  the  tables  give,  the  mild 
structural  and  merchant  steels  may  have  values  almost  as  low 


ART.  80. 


MEAN   CONSTANTS. 


as  the  average  constants  for  wrought  iron.  In  general,  there- 
fore,  the  following  values  should  not  be  used  in  actual  cases  of 
investigation  and  design  except  for  approximate  computations. 

Detailed  tables  giving  the  results  of  experiments  upon 
numerous  kinds  and  qualities  of  materials  may  be  found  in  the 
various  engineers'  pocket  books,  in  THURSTON'S  Materials  of 
Engineering  (New  York,  1884),  in  BURR'S  Elasticity  and 
Strength  of  Materials  (New  York,  1888),  in  JOHNSON'S 
Materials  of  Construction  (New  York,  1897),  and  in  the  works 
noted  at  the  end  of  the  next  article.  It  is,  however,  impos- 
sible to  ascertain  the  exact  strength  of  any  particular  lot  of 
material  by  reference  to  books,  but  tests  must  be  made. 

TABLE  I. 


Mean  Weight. 

Coefficient  of  Linear  Expansion. 

Material. 

Pounds  per 
cubic  foot. 

Kilograms  per 
cubic  meter. 

For  i°  Fah. 

For  i°  Cent. 

Timber, 

40 

600 

0.0000020 

0.0000036 

Brick, 

125 

2  000 

0.0000050 

0.0000090 

Stone, 

160 

2  560 

0.0000050 

0.0000090 

Cast  Iron, 

450 

7  200 

0.0000062 

O.OOOOII2 

Wrought  Iron, 

480 

7  700 

0.0000067 

O.OOOOI2I 

Steel, 

490 

7  800 

0.0000065 

O.OOOOII7 

TABLE  II. 


Material. 

Elastic  Limit. 

Coefficient  of  Elasticity. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms  per 
square  centimeter. 

Timber, 

3  000 

210 

I  500000 

105  ooo 

Cast  Iron, 

6  000 

420 

15  000000 

i  050  ooo 

Wrought  Iron, 

25  ooo 

1750 

25  oooooo 

I  750000 

Steel, 

50000 

3500 

30000000 

2  100  000 

104. 


THE   STRENGTH    OF   MATERIALS 

TABLE  III. 


CH.  VIII 


Material. 

Ultimate  Tensile  Strength. 

Ultimate  Compressive  Strength. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Timber, 

10  000 

700 

8  000 

560 

Brick, 

200 

14 

2  500 

175 

Stone, 

6  000 

420 

Cast  Iron, 

2O  OOO 

I  400 

go  ooo 

6  300 

Wrought  Iron, 

55  ooo 

3850 

55  ooo 

3850 

Steel, 

100  000 

7  ooo 

150  ooo 

10  500 

TABLE  IV. 


Ultimate  Shearing  Strength. 

Modulus  of  Rupture. 

Material. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Timber, 

(          600    ) 
(       3  ooo    f 

\            42    I 

(             210     ) 

9  ooo 

630 

Stone, 

2  000 

140 

Cast  Iron, 

20  000 

I  4OO 

35  ooo 

2  450 

Wrought  Iron, 

50  ooo 

3  500 

Steel, 

70000 

4900 

ART.  81.    HISTORICAL. 

Some  topics  in  the  mechanics  of  materials  were  discussed 
and  partially  developed  in  advance  of  precise  knowledge  re- 
garding elastic  resistance  and  ultimate  strength.  The  first 
investigations  were  those  by  GALILEI  in  1638  on  the  flexure 
of  beams  and  on  forms  of  uniform  strength.  In  1678  HOOKE 
announced  the  law  "  ut  tensio  sic  vis,"  namely,  that  the  elon- 
gation of  a  spring  is  proportional  to  the  force  which  causes  it ; 
three  years  earlier  he  had  performed  experiments  in  the  pres- 
ence of  the  king  of  England  illustrating  the  law,  which  indeed 


ART.  81.  HISTORICAL.  165 

he     had     published     in    1660    concealed    in    the     anagram 
"  ceiilnosssttu  v." 

During  the  eighteenth  century  but  slight  advances  were 
made  in  practical  knowledge,  although  the  theory  of  flexure 
was  improved  by  BERNOULLI,  LEIBNITZ,  COULOMB,  EULER 
and  others.  The  few  experimental  results  during  this  century 
were  on  the  rupture  of  timber  by  flexure  and  by  tension,  ques- 
tions of  ultimate  strength  being  only  investigated  while  that  of 
elastic  limit  was  scarcely  recognized. 

Early  in  the  nineteenth  century  appeared  the  *  Lectures  on 
Natural  Philosophy '  by  YOUNG  in  which  is  found  a  clear  pres- 
entation of  many  of  the  laws  of  flexure  both  under  static 
forces  and  under  shock.  It  also  introduces  for  the  first  time 
the  coefficient  or  modulus  of  elasticity,  £,  but  fails  to  note  that 
it  can  only  be  deduced  or  applied  when  the  elastic  limit  of  the 
material  is  not  surpassed.  A  little  later  BARLOW,  TREDGOLD, 
and  HODGKINSON  experimented  on  timber  and  cast  iron,  both 
in  the  form  of  beams  and  of  columns  ;  their  methods  and  re- 
sults, although  now  seeming  rude  and  defective,  are  deserving 
of  praise  as  the  first  of  real  practical  value. 

The  complete  theory  of  the  flexure  of  beams  and  the  equa- 
tions of  the  elastic  curve  are  due  to  NAVIER,  who  from  1820 
to  1833  established  the  modern  mathematical  theory  of  elas- 
ticity on  a  solid  foundation,  which  by  his  followers  LAME,  ST. 
VENANT  and  BOUSSINESQ  has  been  applied  to  very  complex 
problems,  a  full  account  of  which  is  given  in  TODHUNTER  and 
PIERSON'S  History  of  the  Elasticity  and  Strength  of  Materials 
(3  volumes,  1886-1892). 

In  1849  was  published  the  '  Report  of  the  Commissioners  on 
the  Application  of  Iron  to  Railway  Structures/  which  may  be 
regarded  as  the  landmark  of  the  beginning  of  modern  methods 
of  testing.  It  contains  the  records  of  valuable  tests  by  WILLIS, 
JAMES,  HODGKINSON  and  GALTON  on  the  strength  of  cast 


l66  THE  STRENGTH  OF  MATERIALS.  CH.  VIII. 

and  wrought  iron  as  well  as  upon  the  resistance  to  impact, 
investigations  of  the  increase  in  stress  caused  by  a  rolling  load 
on  a  beam,  experiments  on  the  fatigue  of  metals,  and  the  evi- 
dence given  by  leading  British  engineers  as  to  their  opinions 
on  proper  factors  of  safety  under  different  conditions.  The 
immediate  result  of  this  report  was  the  decision  by  the  English 
board  of  trade  that  the  factor  of  safety  for  cast,  iron  should 
be  twice  as  great  for  rolling  loads  as  for  steady  ones,  while 
throughout  both  Europe  and  the  United  States  it  excited  a 
marked  interest  and  impetus  in  the  subject  of  testing  materials. 

A  volume  would  be  required  to  outline  the  progress  and  the 
results  of  the  experimental  work  done  since  1850.  The  main 
conclusions  will  be  noted  in  subsequent  articles,  and  many 
others  will  be  found  in  the  books  noted  in  Art.  80.  The  fol- 
lowing additional  references  to  works  of  the  principal  experi- 
menters will  enable  students  to  consult  original  authorities  as 
well.  It  should  be  noted,  however,  that  very  important  contri- 
butions have  appeared  in  technical  periodicals  and  in  the  trans- 
actions of  engineering  societies  ;  for  the  titles  of  many  of  these 
see  *  Descriptive  Index  of  Engineering  Literature,'  published 
in  1892,  1896,  and  1901. 

WADE,  W.,  and  RODMAN,  T.  J.:  Reports  of  Experiments 
on  the  Strength  and  other  Properties  of  Metals  for  Cannon. 
Two  volumes,  quarto:  Philadelphia,  1856,  pp.  482;  Boston, 
1 86 1,  pp.  308. 

KIRKALDY,  D.:  Experiments  on  Wrought  Iron  and  Steel. 
Second  edition,  London,  1861,  octavo,  pp.  227,  plates  xvi. 

FAIRBAIRN,  W.:  An  Experimental  Inquiry  into  the 
Strength,  Elasticity,  Ductility  and  other  Properties  of  Steel. 
London,  1869,  octavo,  pp.  51. 

STYFFE,  K.:  The  Elasticity,  Extensibility  and  Tensile 
Strength  of  Iron  and  Steel.  London,  1869,  octavo,  pp.  171, 
plates  ix. 

BAUSCHINGER,   J.:     Mittheilungen    aus    dem    mechanisch- 


ART.  82.  TESTING  MACHINES.  l/ 

technischen  Laboratorium  der  polytechnischen  Schule  in 
Munchen.  Munich,  1873-1890,  folio;  usually  published  annu- 
ally, each  part  dealing  with  a  special  subject. 

GILMORE,  Q.  A.:  The  Compressive  Resistance  of  Free- 
stone, Brick  Piers,  Hydraulic  Cements,  Mortars  and  Concretes. 
New  York,  1888,  octavo,  pp.  198,  plates  viii. 

Reports  of  the  U.  S.  Board  appointed  to  Test  Iron,  Steel, 
and  other  Metals.  Washington,  two  volumes,  1878  and  1881, 
octavo,  pp.  592  and  684,  with  many  plates. 

Bulletins  of  the  Forestry  Division  of  the  U.  S.  Department 
of  Agriculture  since  1892.  Also,  Report  of  Tenth  Census  on 
Forest  Trees  of  North  America. 

U.  S.  Ordnance  Bureau  :  Tests  of  Metals.  Annual  records 
of  tests  at  Watertown  Arsenal  since  1883. 

Baumaterialienkunde.  A  semi-monthly  journal  published 
at  Stuttgart  since  1896;  the  official  organ  of  the  International 
Association  for  Testing  Materials. 

Prob.  134.  Consult  Engineering  News,  Aug.  17,  1899,  and 
ascertain  the  history  and  objects  of  the  International  Associa- 
tion for  Testing  Materials. 

ART.  82.    TESTING  MACHINES. 

Tests  on  the  flexure  and  rupture  of  beams  were  made  in  the 
seventeenth  century,  and  machines  for  this  purpose  are  com- 
paratively simple.  The  load  is  usually  applied  at  the  middle 
of  a  beam  supported  at  its  ends  and  gradually  increased  until 
rupture  occurs,  the  deflection  being  measured  also  as  a  test  of 
stiffness.  The  apparatus  for  determining  the  deflection  should 
be  attached  to  the  supports  so  that  the  compression  of  these 
may  not  affect  the  observed  values.  In  the  simplest  case 
weights  are  hung  upon  a  ring  or  stirrup  placed  upon  the  mid- 
dle of  the  beam,  and  are  added  in  regular  increments  of  100 
pounds,  more  or  less,  depending  upon  the  size  of  the  beam. 
When  the  elastic  limit  is  not  exceeded  the  coefficient  of  elas- 
ticity E  may  be  computed  from  the  observed  deflection  by  the 


l68  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

formula  in  Case  I  of  Art.  35.  From  the  breaking  load  P  the 
modulus  of  rupture  Sr  may  be  computed  by  the  formula  (4) 
of  Art.  21  ;  the  value  of  this  lies  between  the  ultimate  tensile 
and  compressive  strengths  of  the  material.  As  a  method  of 
comparison  of  different  qualities  of  materials  this  test  is  an 
excellent  one. 

Machines  for  tensile  tests  are  usually  arranged  so  as  to 
operate  on  a  specimen  of  the  shape  shown  in  Fig.  2  of  Art.  8. 
The  heads  are  either  griped  in  jaws  or  are  provided  with 
threads  so  that  they  may  be  screwed  into  nuts  to  which  the 
tension  is  applied.  The  power  may  be  furnished  by  a  lever  in 
machines  of  small  capacity,  in  others  by  a  screw  or  by  hydrau- 
lic pressure,  the  total  tension  being  weighed  off  on  a  compound 
lever.  In  commercial  tests  the  ultimate  elongation  is  alone 
measured  ;  this  is  done  by  marking  inch  spaces  on  the  speci- 
men and  measuring  them  after  rupture.  In  scientific  tests  an 
extensimeter  is  attached  to  the  specimen  so  that  the  elonga- 
tion can  be  read  at  each  increment  of  weight.  The  elongation 
is  usually  expressed  as  a  percentage  of  the  original  length 
between  two  marks  whose  distance  apart  is  more  than  eight 
times  the  diameter  of  the  specimen.  In  the  case  of  ductile 
metals  a  marked  diminution  in  diameter  occurs  at  the  point  of 
rupture,  and  the  two  parts  of  the  specimen  are  seen  to  have  a 
taper  on  each  side  of  the  fracture.  On  this  account  the  per- 
centage of  ultimate  elongation  will  depend  upon  the  distance 
between  the  two  marks.  As  no  standard  proportions  have  yet 
been  adopted,  it  is  desirable  that  the  actual  distance  on  which 
the  elongation  is  measured  should  be  always  stated. 

In  ductile  materials,  like  wrought  iron  and  mild  steel,  the 
final  strength  is  less  than  the  maximum  strength,  owing  to  the 
rapid  flow  of  metal  which  is  the  cause  of  the  taper  and  con- 
traction. The  maximum  strength  is  usually  alone  recorded, 
and  in  this  volume  the  term  '  ultimate  strength '  is  to  be  un- 


ART.  82.  TESTING   MACHINES.  169 

derstood,  not  as  that  at  the  instant  of  rupture,  but  as  the  maxi- 
mum unit-stress  observed  shortly  before  rupture. 

The  contraction  of  area,  introduced  by  KlRKALDY  as  an 
element  to  be  noted  in  tests  of  ductile  metals,  is  now  regarded 
as  of  equal  importance  with  the  ultimate  elongation,  since  it  is 
subject  to  less  variation  with  the  length  of  the  specimen.  This 
is  expressed  as  a  percentage  of  the  original  area  of  the  cross- 
section.  According  to  KlRKALDY,  the  tensile  strength  and 
ultimate  contraction  of  area,  when  considered  jointly,  furnish 
the  best  index  for  judging  the  quality  of  wrought  iron  and 
steel. 

In  all  these  tensile  tests  the  load  is  applied  gradually,  and 
not  suddenly  or  with  impact.  A  test,  however,  may  be  made 
slowly  or  quickly,  and  it  is  found  that  the  degree  of  rapid- 
ity has  a  marked  influence  upon  the  results.  In  general,  a 
quick  test  gives  a  higher  elastic  limit  and  a  higher  strength 
than  a  slow  one,  while  the  ultimate  elongation  is  less.  Attempts 
have  hence  been  made  to  specify  the  degree  of  rapidity  with 
which  the  test  should  be  conducted,  and  undoubtedly  some 
uniformity  in  this  respect  will  in  time  be  required  in  standard 
specifications.  (Art.  88.) 

Compressive  tests  are  mainly  confined  to  brick  and  stone, 
and  are  but  little  used  for  metals  on  account  of  the  difficulty 
of  securing  a  uniform  distribution  of  stress  over  the  surfaces. 
Rupture  in  these  cases  rarely  occurs  by  true  crushing,  but  by 
a  diagonal  shearing  or  splitting,  and  it  is  not  easy  to  obtain 
precise  measures  of  the  amount  of  shortening.  Cement,  which 
is  always  used  in  compression;  is  indeed  usually  tested  by  ten- 
sion, and  this  is  found  to  be  the  cheaper  and  more  satisfactory 
method. 

The  first  testing  machines  in  the  United  States  were  those 
built  by  WADE  and  RODMAN  between  1850  and  1860  for  test- 
ing gun-metal  for  the  government.  About  this  time  the 


1/0  THE  STRENGTH   OF   MATERIALS.  CH.  VIIL 

rapid  introduction  of  iron  bridges  led  to  experiments  by 
PLYMPTON  and  by  ROEBLING.  About  1865  machines  were 
made  by  FAIRBANKS,  which  have  since  been  developed  into 
forms  applicable  to  all  classes  of  work,  as  also  have  those  made 
by  OLSEN  and  by  RIEHLE.  A  machine  devised  by  THURSTON 
soon  after  1870  has  done  excellent  work  on  stresses  of  torsion. 
The  machine  built  by  EMERY  for  the  United  States  Board  of 
1878-81,  and  now  in  the  Watertown  arsenal,  is  the  most  pre- 
cise one  ever  devised,  and  its  work  has  been  of  great  value  in 
advancing  our  knowledge  of  materials.  Large  machines  for 
testing  eye-bars  have  been  built  by  bridge  companies  since 
1880,  and  several  testing  laboratories  now  exist  provided  with 
machines  for  every  kind  of  work. 

The  capacity  of  a  testing  machine  is  the  tension  or  pressure 
which  it  can  exert.  A  small  machine  for  testing  cement  need 
not  have  a  capacity  greater  than  1000  pounds.  Machines  of 
50000,  looooo,  and  150000  pounds  for  testing  metals  are  com- 
mon. The  Watertown  machine  has  a  capacity  of  I  ooo  ooo 
pounds  and  can  test  a  heavy  bar  30  feet  long  and  a  small  hair 
with  equal  precision.  The  eye-bar  machine  at  Athens,  Pa., 
has  a  capacity  of  I  244  ooo  pounds,  and  that  at  Phcenixville, 
Pa.,  a  capacity  of  2  160000  pounds. 

For  descriptions  of  the  principal  testing  machines  the  stu- 
dent should  consult  ABBOTT'S  papers  in  Van  Nostrand's 
Magazine,  Vol.  XXX,  reprinted  as  Van  Nostrand's  Science 
Series,  No.  74.  KENT'S  Strength  of  Materials,  No.  41  in  that 
Series,  contains  also  valuable  information  and  discussions  of 
methods  of  testing.  JOHNSON'S  Materials  of  Construction, 
1897,  and  MARTENS'  Handbook  of  Testing  Materials,  .1899, 
treat  fully  of  testing  machines  and  of  the  methods  of  making 
tests  so  that  the  results  shall  be  reliable. 

Prob.  135.  A  steel  eye-bar,  10  X  2f  inches,  tested  at  Phce- 
nixville in  1 893,  broke  under  a  tension  of  I  626  322  pounds,  with 


ART.  83. 


TIMBER. 


171 


20.5  per  cent  elongation  in  47  feet,  and  a  reduction  of  area  of 
50.4  per  cent.  Compute  the  ultimate  tensile  strength  per 
square  inch  of  original  area,  and  also  per  square  inch  of  frac- 
tured area. 

ART.  83.    TIMBER. 

Good  timber  is  of  uniform  color  and  texture,  free  from 
knots,  sapwood,  wind-shakes,  worm-holes  or  decay ;  it  should 
also  be  well  seasoned,  which  is  best  done  by  exposing  it  for 
two  or  three  years  to  the  weather  to  dry  out  the  sap.  The 
heaviest  timber  is  usually  the  strongest  ;  also  the  darker  the 
color  and  the  closer  the  annular  rings  the  stronger  and  better 
it  is,  other  things  being  equal.  The  strength  of  timber  is  al- 
ways greatest  in  the  direction  of  the  grain,  the  sidewise  resist- 
ance to  tension  or  compression  being  scarcely  one  fourth  of 
the  longitudinal. 

The  following  table  gives  average  values  of  the  ultimate 
strength  in  pounds  per  square  inch  of  a  few  of  the  common 


Kind. 

Pounds 
per 
Cubic  Foot. 

Tensile 
Strength. 

Modulus 
of 
Rupture. 

Compress- 
ive 
Strength. 

Hemlock  

2C 

8  ooo 

6000 

5  ooo 

White  Pine  

27 

AQ 

8000 

12  OOO 

60OO 
7  OOO 

5500 

e  OOO 

Red  Oak  

42 

Q  OOO 

7  OOO 

6  OOO 

Yellow  Pine... 
White  Oak  

45 

48 

I50OO 
12  OOO 

IIOOO 
10  OOO 

9OOO 
8000 

kinds  of  timber  as  determined  from  tests  of  small  specimen 
carefully  selected  and  dried.  Large  pieces  of  timber  such  ai 
are  actually  used  in  engineering  structures  will  probably  have 
an  ultimate  strength  of  from  fifty  to  eighty  per  cent  of  these 
values.  Moreover  the  figures  are  liable  to  a  range  of  25  per 
cent  on  account  of  variations  in  quality  and  condition  arising 
from  place  of  growth,  time  when  cut,  and  method  and  duration 


172  THE  STRENGTH   OF  MATERIALS.  Ctl.  VIII. 

of  seasoning.    To  cover  these  variations  the  factor  of  safety  of 
10  is  not  too  high. 

The  shearing  strength  of  timber  is  still  more  variable  than 
the  tensile  or  compressive  resistance.  White  pine  across  the 
grain  may  be  put  at  2500  pounds  per  square  inch,  and  along 
the  grain  at  500.  Chestnut  has  1500  and  600  respectively, 
yellow  pine  and  oak  perhaps  4  ooo  and  600  respectively. 

The  elastic  limit  of  timber  is  poorly  defined.  In  precise 
tests  on  good  specimens  it  is  sometimes  observed  at  about  one 
half  the  ultimate  strength,  but  under  ordinary  conditions  it  is 
safer  to  put  it  at  one  third.  The  coefficient  of  elasticity 
ranges  from  i  ooo  ooo  to  2  ooo  ooo  pounds  per  square  inch, 
I  500  ooo  being  a  good  mean  value  to  use  in  general  computa- 
tions. The  ultimate  elongation  is  small,  usually  being  between 
I  and  2  per  cent. 

The  tests  published  in  the  Census  Report  on  the  Forest 
Trees  of  North  America  (1884)  are  very  comprehensive  as  they 
include  412  species  of  timber.  Of  these  16  species  have  a 
specific  gravity  greater  than  i.o  and  28  species  less  than  0.4. 
Even  in  the  same  species  a  great  variation  in  weight  was  often 
found  ;  for  instance  white  oak  ranged  from  42  pounds  to  54 
pounds  per  cubic  foot.  The  heaviest  wood  weighed  81  pounds 
per  cubic  foot  and  had  a  compressive  strength  of  12  ooo  pounds 
per  square  inch;  the  lightest  wood  weighed  16  pounds  per 
cubic  foot  and  had  a  compressive  strength  of  200  pounds  per 
square  inch. 

ART.  84.    BRICK. 

Brick  is  made  of  clay  which  consists  mainly  of  silicate  of 
alumina  with  compounds  of  lime,  magnesia,  and  iron.  The  clay 
is  prepared  by  cleaning  it  carefully  from  pebbles  and  sand, 
mixing  it  with  about  one  half  its  volume  of  water,  a'nd  temper- 
ing it  by  hand  stirring  or  in  a  pug-mill.  It  is  then  moulded  in 


ART.  84.  BRICK.  173 

rectangular  boxes  by  hand  or  by  special  machines,  and  the 
green  bricks  are  placed  under  open  sheds  to  dry.  These  are 
piled  in  a  kiln  and  heated  for  nearly  two  weeks  until  those 
nearest  to  the  fuel  assume  a  partially  vitrified  appearance. 

Three  qualities  of  brick  are  taken  from  the  kiln ;  '  arch 
brick  '  are  those  from  around  the  arches  where  the  fuel  is 
burned — these  are  hard  and  often  brittle  ;  *  body  brick,'  from 
the  interior  of  the  kiln,  are  of  the  best  quality ;  '  soft  brick,' 
from  the  exterior  of  the  pile,  are  weak  and  only  suitable  for 
filling.  Paving  brick  are  burned  in  special  kilns,  often  by 
natural  gas  or  by  oil,  the  rate  of  heating  being  such  as  to  en- 
sure toughness  and  hardness. 

The  common  size  is  2  X  4  X  8J  inches,  and  the  average 
weight  4^  pounds.  A  pressed  brick,  however,  may  weigh 
nearly  5^  pounds.  Good  bricks  should  be  of  regular  shape, 
have  parallel  and  plane  faces,  with  sharp  angles  and  edges. 
They  should  be  of  uniform  texture  and  when  struck  a  quick 
blow  should  give  a  sharp  metallic  ring.  The  heavier  the 
brick,  other  things  being  equal,  the  stronger  and  better  it  is. 

Poor  brick  will  absorb  when  dry  from  20  to  30  per  cent  of 
its  weight  of  water,  ordinary  qualities  absorb  from  10  to  20 
per  cent,  while  hard  paving  brick  should  not  absorb  more  than 
2  or  3  per  cent.  An  absorption  test  is  valuable  in  measuring 
the  capacity  of  brick  to  resist  the  disintegrating  action  of 
frost,  and  as  a  rough  general  rule  the  greater  the  amount  of 
water  absorbed  the  less  is  the  strength  and  durability. 

The  crushing  strength  of  brick  is  variable  ;  while  a  mean 
value  may  be  2500  pounds  per  square  inch,  soft  brick  will 
scarcely  stand  500,  pressed  brick  may  run  to  10  ooo,  and  the 
best  qualities  of  paving  brick  have  given  15000  pounds  per 
square  inch,  or  even  more.  Crushing  tests  are  difficult  and  ex- 
pensive to  make  on  account  of  the  labor  of  preparing  the  speci- 
men cubes  so  as  to  secure  surfaces  truly  parallel. 


1/4  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

A  flexural  test  for  brick  is  often  used.  The  brick  is  sup- 
ported at  the  middle  of  the  flat  side  upon  a  fixed  steel  support 
about  one  half  inch  in  width,  and  around  its  ends  are  placed 
stirrups  to  which  is  hung  a  vessel  for  holding  weights.  The 
loads  being  applied  until  rupture  occurs,  the  modulus  of  rup- 
ture can  be  computed  from  the  formula 


in  which  /  is  the  length  of  the  brick  between  stirrups,  b  its 
breadth,  d  its  thickness,  and  W  the  breaking  load.  For  ex- 
ample, if  /  =  8  inches,  b  —  4  inches,  d  —  2  inches,  and  W  = 
1000  pounds,  then  Sr  =  750  pounds  per  square  inch.  This 
modulus  of  rupture  is  intermediate  between  the  ultimate  ten- 
sile and  compressive  strengths,  and  although  not  a  physical 
constant  it  serves  the  means  of  making  excellent  comparisons 
of  different  qualities  of  brick. 

Tensile  and  shearing  tests  of  bricks  are  rarely  made  and  but 
little  is  known  of  their  behavior  under  such  stresses.  The  ulti- 
mate tensile  strength  may  perhaps  range  from  50  to  500  pounds 
per  square  inch.  As  for  elastic  limit  and  coefficient  of  elas- 
ticity, so  little  is  known  that  nothing  can  be  said. 

ART.  85.    CEMENT  AND  MORTAR. 

Common  mortar  is  composed  of  one  part  of  lime  to  five 
parts  of  sand  by  measure.  When  six  months  old  its  tensile 
strength  is  from  15  to  30,  and  its  compressive  strength  from 
150  to  300  pounds  per  square  inch.  Its  strength  slowly  in- 
creases with  age,  and  it  may  be  considerably  increased  by  using 
a  smaller  proportion  of  sand. 

Hydraulic  mortar  is  composed  of  hydraulic  cement  and 
sand  in  varying  proportions.  The  less  the  proportion  of  sand 
the  greater  is  its  strength.  If  5  be  the  strength  of  neat  cement, 
that  is,  cement  with  no  sand,  and  Sl  be  the  strength  of  mortar 


ART.  85. 


CEMENT   AND   MORTAR. 


175 


having/  parts  of  sand  to  i  part  of  cement,  the  formula 

5  -      5 

gives  a  rough  approximation  to  the  strength  of  the  mortar. 
A  common  proportion  is  3  parts  of  sand  to  I  of  cement,  the 
strength  of  this  being  about  one-fourth  that  of  the  cement  it- 
self. The  strength  of  hydraulic  mortar  also  increases  with 
its  age. 

There  are  two  classes  of  hydraulic  cements,  the  Rosendale 
or  natural  cements,  and  the  Portland  or  artificial  cements. 
The  former  are  of  lighter  color,  lower  weight,  and  lesser 
strength  than  the  latter,  but  they  are  quicker  in  setting  and 
cheaper  in  price.  The  following  table  gives  average  ultimate 
tensile  strengths  in  pounds  per  square  inch  of  mortars  of  both 
classes  of  different  ages  and  different  proportions  of  sand. 


Propor- 
tion of 
Sand. 

ROSENDALE. 

PORTLAND. 

One 
Week. 

One 

Month. 

Six 

Months. 

One 
Year. 

One 
Week. 

One 
Month. 

Six 

Months. 

One 
Year. 

O 

1  2O 

200 

300 

350 

300 

400 

450 

500 

I  tO  I 

70 

125 

200 

250 

125 

200 

300 

350 

2  tO  I 

30 

60 

120 

180 

100 

ISO 

250 

300 

3  to  i 

20 

30 

70 

130 

80 

«5 

200 

250 

4  to  i 

10 

2O 

60 

100 

50 

70 

120 

150 

These  figures  are  obtained  from  tests  of  briquettes  carefully 
made  in  molds,  and  are  probably  higher  than  mortar  made 
under  usual  circumstances  would  give.  The  briquette  is  re- 
moved from  the  mold  when  set,  allowed  to  remain  in  air  for 
one  day,  and  then  put  under  water  for  the  remainder  of  the 
time. 

As  cements  and  mortars  are  only  used  in  compression,  the 
tensile  test  may  seem  an  inappropriate  one.  Compressive  tests, 
however,  are  expensive  and  unsatisfactory  to  make  under  or- 


176  THE   STRENGTH    OF   MATERIALS.  CH.  VIII. 

dinary  conditions.  It  may  be  taken  as  a  general  rule  that  the 
compressive  strength  of  a  material  increases  with  the  tensile 
strength,  and  it  is  certain  that  the  universal  adoption  of  the 
tensile  tests  has  done  much  to  greatly  improve  the  quality  of 
hydraulic  cements. 

For  neat  cement  a  one-day  test  is  frequently  employed, 
the  briquette  being  put  under  water  as  soon  as  it  has  set.  For 
Rosendale  cement  a  briquette  one  square  inch  in  section  should 
give  a  tensile  strength  of  75  pounds,  while  one  of  Portland 
cement  should  give  125  pounds.  To  secure  these  results,  how- 
ever, the  material  should  be  thoroughly  rammed  into  the 
moulds,  no  more  water  being  used  than  necessary,  and  the 
quality  of  the  cement  must  be  good.  . 

The  compressive  strength  of  hydraulic  cements  and  mortars 
is  much  higher  than  the  tensile  strength,  and  it  increases  more 
rapidly  with  age.  A  common  statement  is  that  the  compressive 
strength  is  from  eight  to  ten  times  the  tensile  strength.  Neat 
Portland  cement  when  one  month  old  has  a  compressive 
strength  of  about  3  ooo,  and  when  one  year  old  about  5  ooo 
pounds  per  square  inch.  Rosendale  cement  mortar  when  three 
or  four  years  old  has  about  2  500  pounds  per  square  inch. 

The  adhesion  of  cement  to  stone  or  brick  is  somewhat  less 
than  the  tensile  strength.  The  shearing  strength  of  cement  or 
mortar  is  much  less  than  the  tensile  strength — usually  only 
about  one-fourth. 

The  strength  of  cement  and  mortar  is  influenced  by  many 
causes  :  the  quality  of  the  stone  or  materials  from  which  the 
cement  is  made,  the  method  of  manufacture,  the  age  of  the 
cement,  the  kind  of  sand,  the  method  of  mixing,  and  even  the 
amount  of  water  used.  Tensile  tests  of  briquettes,  more  than 
all  others,  are  valuable  on  account  of  the  ease  with  which  they 
can  be  made,  and  the  reliable  conclusions  that  can  be  drawn 
from  the  results. 


ART.  86. 


STONE. 


177 


ART.  86.    STONE. 

Sandstone,  as  its  name  implies,  is  sand,  usually  quartzite, 
which  has  been  consolidated  under  heat  and  pressure.  It 
varies  much  in  color,  strength,  and  durability,  but  many  varie- 
ties form  most  valuable  building  material.  In  general  it  is 
easy  to  cut  and  dress,  but  the  variety  known  as  Potsdam  sand- 
stone is  very  hard  in  some  localities. 

Limestone  is  formed  by  consolidated  marine  shells,  and  is 
of  diverse  quality.  Marble  is  limestone  which  has  been  re- 
worked in  the  laboratory  of  nature  so  as  to  expel  the  impuri- 
ties, and  leave  a  nearly  pure  carbonate  of  lime  ;  it  takes  a  high 
polish,  is  easily  worked,  and  makes  one  of  the  most  beautiful 
building  stones. 

Granite  is  a  rock  of  aqueous  origin  metamorphosed  under 
heat  and  pressure ;  its  composition  is  quartz,  feldspar,  and 
mica,  but  in  the  variety  called  syenite  the  mica  is  replaced  by 
hornblende.  It  is  fairly  easy  to  work,  usually  strong  and  dur- 
able, and  some  varieties  will  take  a  high  polish. 

Trap,  or  basalt,  is  an  igneous  rock  without  cleavage.  It  is 
hard  and  tough,  and  less  suitable  for  building  constructions 
than  other  rocks,  as  large  blocks  cannot  be  readily  obtained 
and  cut  to  size.  It  has,  however,  a  high  strength,  and  is  re- 
markable for  durability. 

The  average  weights  and  ultimate  compressive  strengths  of 
these  four  classes  are  as  follows  : 


Kind. 

Weight 
per  Cubic 
Foot. 

Compress- 
ive 
Strength. 

Modulus 
of 
Rupture. 

Sandstone 

Pounds. 
150 

Lbs.  per 
sq.  in. 
5000 

Lbs.  per 
sq.  in. 
I  500 

Limestone 

1  60 

7000 

I  5OO 

Granite 

I65 

12000 

2000 

Trap 

175 

I6OOO 

1^8  THE   STRENGTH   OF   MATERIALS.  ClI.  VIII. 

These  figures,  however,  refer  to  small  specimens  such  as 
can  be  used  in  a  testing  machine,  and  it  is  known  that  the 
strength  of  large  blocks  per  square  inch  is  materially  less.  The 
rupture  of  a  cube  or  prism  of  stone  under  compression  often 
occurs  by  splitting,  or  rather  shearing,  in  planes  making  an 
angle  of  about  45  degrees  with  the  direction  of  the  pressure 
(see  Fig.  3,  Art.  8).  In  order  to  insure  a  perfect  distribution 
of  pressure  over  the  surfaces,  cushions  of  wood,  leather,  and 
lead  are  often  placed  upon  them,  but  the  advantage  of  using 
them  is  doubtful. 

The  coefficient  of  elasticity  of  stone  has  been  found  to 
range  from  5000000  to  10000000  pounds  per  square  inch. 
The  elastic  limit  is  difficult  to  observe,  if  indeed  any  exists. 
Little  is  known  of  .the  tensile  and  shearing  strengths  of  stone, 
except  that  they  are  smaller  than  the  compressive  strength  in 
some  varieties.  Tests  to  determine  the  modulus  of  rupture  by 
loading  a  stone  beam  to  destruction  are  easily  made,  and  will 
probably  serve  to  compare  the  quality  of  different  specimens 
better  than  other  tests  of  strength. 

Slate  is  an  argillaceous  stone  consolidated  under  very  heavy 
pressure  so  as  to  form  a  marked  cleavage  at  right  angles  to  the 
direction  of  that  pressure.  It  is  split  into  plates  J  inch  in 
thickness  for  use  as  roofing  slate,  and  in  larger  blocks  is  used 
for  pavements  and  steps.  Its  weight  per  cubic  foot  is  about 
175  pounds,  its  compressive  strength  about  10000  pounds  per 
square  inch,  and  its  modulus  of  rupture  about  7  ooo  pounds 
per  square  inch.  Unfortunately  it  is  liable  to  corrode  under 
the  action  of  the  atmosphere,  and  its  marked  cleavage  and 
grain  render  its  strength  variable  in  different  directions.  See 
Transactions  American  Society  of  Civil  Engineers,  Sept.  1892 
and  Dec.  1894. 

The  quality  of  a  building  stone  cannot  be  safely  inferred 
from  tests  of  strength,  as  its  durability  depends  largely  upon 


ART.  87.  CAST  IRON.  179 

its  capacity  to  resist  the  action  of  the  weather.  Hence  cor- 
rosion and  freezing  tests,  impact  tests,  and  observations  of  the 
behavior  of  stone  under  conditions  of  actual  use  are  more  im- 
portant than  the  determination  of  crushing  strength  in  a  com- 
pression machine.  See  BAKER'S  Masonry  Construction  for 
full  information  regarding  these  subjects. 

ART.  87.    CAST  IRON. 

Cast  iron  is  a  modern  product,  having  been  first  made  in 
England  about  the  beginning  of  the  fifteenth  century.  Ores 
of  iron  are  melted  in  a  blast  furnace,  producing  pig  iron.  The 
pig  iron  is  remelted  in  a  cupola  furnace  and  poured  into 
moulds,  thus  forming  castings.  Beams,  columns,  pipes,  braces, 
and  blocks  of  every  shape  required  in  engineering  structures 
are  thus  produced. 

Pig  iron  is  divided  into  two  classes,  Foundry  pig  and  Forge 
pig,  the  former  being  used  for  castings  and  the  latter  for  mak- 
ing wrought  iron.  Foundry  pig  has  a  dark-gray  fracture,  with 
large  crystals  and  a  metallic  luster ;  forge  pig  has  a  light-gray 
or  silver-white  fracture,  with  small  crystals.  Foundry  pig  has 
a  specific  gravity  of  from  7.1  to  7.2,  and  it  contains  from  6  to 
4  per  cent  of  carbon ;  forge  pig  has  a  specific  gravity  of  from 
7.1  to  7.4,  and  it  contains  from  4  to  2  per  cent  of  carbon.  The 
higher  the  percentage  of  carbon  the  less  is  the  specific  gravity, 
and  the  easier  it  is  to  melt  the  pig.  Besides  the  carbon  there 
are  present  from  i  to  5  per  cent  of  other  impurities,  such  as 
silicon,  manganese,  and  phosphorus. 

The  properties  and  strength  of  castings  depend  upon  the 
quality  of  the  ores  and  the  method  of  their  manufacture  in 
both  the  blast  and  the  cupola  furnace.  Cold-blast  pig  produces 
stronger  iron  than  the  hot-blast,  but  it  is  more  expensive. 
Long-continued  fusion  improves  the  quality  of  the  product,  as 
also  do  repeated  meltings.  The  darkest  grades  of  foundry  pig 


ISO  THE  STRENGTH   OF   MATERIALS.  CH.  VIII. 

make  the  smoothest  castings,  but  they  are  apt  to  be  brittle  ; 
the  light-gray  grades  make  tough  castings,  but  they  are  apt  to 
contain  blow-holes  or  imperfections. 

The  percentage  of  carbon  in  cast  iron  is  a  controlling  factor 
which  governs  its  strength,  particularly  that  percentage  which 
is  chemically  combined  with  the  iron.  For  example,  the  fol- 
lowing are  the  results  of  tests  by  WADE  of  three  classes  of 
cast  iron  for  guns,  the  tensile  strength  being  expressed  in 
pounds  per  square  inch : 

VT        c       •&    r*       •»  Percentage  of  Carbon.  Ultimate 

No.     Specific  Gravity.        Graphite<  8        Combined.        Tensile  Strength. 

1  7.204  2.06  1.78  28800 

2  7.154  2.30  1.46  24800 

3  7.087  2.83  0.82  20  ioo 

Here  it  is  seen  that  the  total  carbon  is  about  the  same  in  the 
three  kinds,  but  the  smaller  the  percentage  of  combined  carbon 
the  less  is  the  specific  gravity  and  the  ultimate  strength. 

As  average  values  for  the  ultimate  strength  of  cast  iron, 
20000  and  90000  pounds  per  square  inch  in  tension  and  com- 
pression respectively  are  good  figures.  In  any  particular  case, 
however,  a  variation  of  from  10  to  20  per  cent  from  these  values 
may  be  expected,  owing  to  the  great  variation  in  quality. 
For  first-class  gun  iron  WADE  found  a  tensile  strength  of  over 
30000  and  a  compressive  strength  of  over  150000  pounds  per 
square  inch.  On  the  other  hand  medium-quality  castings  often 
have  a  tensile  strength  less  than  16000  pounds  per  square  inch. 

In  tensile  tests  the  elongations  increase  faster  than  in  the 
simple  ratio  of  the  applied  stresses,  so  that  the  elastic  limit  is 
poorly  defined,  and  the  coefficient  of  elasticity  may  range  from 
10  ooo  ooo  to  20  ooo  ooo  pounds  per  square  inch.  HODGKIN- 
SON  deduced  formulas  for  finding  the  elongation  for  different 
stresses,  but  these  are  of  little  practical  importance. 

The  flexural  test  is  a  good  one  for  comparing  the  strength 


ART.  88.  WROUGHT  IRON.  181 

of  different  bars  of  cast  iron.  A  bar  2  X  I  inch  in  cross-section 
and  27  inches  long,  laid  flatwise  on  two  supports  24  inches 
apart,  should  carry  a  load  of  2000  pounds,  or  more,  applied  at 
the  middle  ;  that  is,  the  modulus  of  rupture  should  be  36  ooo 
pounds  per  square  inch,  or  more. 

The  high  compressive  strength  and  cheapness  of  cast  iron 
render  it  a  valuable  material  for  many  purposes,  but  its  brittle- 
ness,  low  tensile  strength,  and  ductility  forbid  its  use  in  struc- 
tures subject  to  variations  of  load  or  to  shocks.  Its  ultimate 
elongation  being  scarcely  one  per  cent,  the  work  required  to 
cause  rupture  in  tension  is  small  compared  to  that  for  wrought 
iron  and  steel,  and  hence  as  a  structural  material  the  use  ot 
cast  iron  is  very  limited.  POLE'S  Iron  as  a  Material  of  Con- 
struction  (London,  1872)  is  an  elementary  work  which  the 
student  may  consult  with  advantage. 

ART.  88.    WROUGHT  IRON. 

The  ancient  peoples  of  Europe  and  Asia  were  acquainted 
with  wrought  iron  and  steel  to  a  limited  extent.  It  is  men- 
tioned in  Genesis,  iv.,  22,  and  in  one  of  the  oldest  pyramids  of 
Egypt  a  piece  of  iron  has  been  found.  It  was  produced,  prob- 
ably, by  the  action  of  a  hot  fire  on  very  pure  ore.  The  ancient 
Britons  built  bloomaries  on  the  tops  of  high  hills,  a  tunnel 
opening  toward  the  north  furnishing  a  draught  for  the  fire 
which  caused  the  carbon  and  other  impurities  to  be  expelled 
from  the  ore,  leaving  behind  nearly  pure  metallic  iron. 

Modern  methods  of  manufacturing  wrought  iron  are 
mainly  by  the  use  of  forge  pig  (Art.  87),  the  one  most  exten- 
sively used  being  the  puddling  process.  Here  the  forge  pig 
is  subjected  to  the  oxidizing  flame  of  a  blast  in  a  reverbera- 
tory  furnace,  where  it  is  formed  into  pasty  balls  by  the  puddler. 
A  ball  taken  from  the  furnace  is  run  through  a  squeezer  to 
expel  the  cinder  and  then  rolled  into  a  muck  bar.  The  muck 


1 82  THE   STRENGTH   OF  MATERIALS.  CH.  VI I L 

bars  are  cut,  laid  in  piles,  heated,  and  rolled,  forming  what  is 
called  merchant  bar.  If  this  is  cut,  piled,  and  rolled  again  a 
better  product  called  best  iron  is  produced.  A  third  rolling 
gives  *  best-best '  iron,  a  superior  quality,  but  high  in  price. 

The  product  of  the  rolling  mill  is  bar  iron,  plate  iron,  shape 
iron,  beams,  and  rails.  Bar  iron  is  round,  square,  and  rectan- 
gular in  section  ;  plate  iron  is  from  i  to  I  inch  thick,  and  of 
varying  widths  and  lengths ;  shape  iron  includes  angles,"  tees, 
channels,  and  other  forms  used  in  structural  work  ;  beams  are 
I  shaped,  and  of  the  deck  or  rail  form  (Art.  31). 

Wrought  iron  is  metallic  iron  containing  less  than  0.25 
per  cent  of  carbon,  and  which  has  been  manufactured  without 
fusion.  Its  tensile  and  compressive  strengths  are  closely  equal 
on  the  average  from  50000  to  55  ooo  pounds  per  square  inch. 
The  elastic  limit  is  well  defined  at  about  25  ooo  pounds  per 
square  inch,  and  within  that  limit  the  law  of  proportionality  of 
stress  to  deformation  is  strictly  observed.  It  is  tough  and 
ductile,  having  an  ultimate  elongation  of  from  20  to  30  per 
cent.  It  is  stiffer  than  cast  iron,  the  coefficient  of  elasticity 
being  25  ooo  ooo  pounds  per  square  inch.  It  is  malleable,  can 
be  forged  and  welded,  and  has  a  high  capacity  to  withstand 
the  action  of  shocks.  It  cannot,  however,  be  tempered  nor 
can  it  be  melted  at  any  ordinary  temperature. 

The  cold-bend  test  for  wrought  iron  is  an  important  one 
for  judging  of  general  quality.  A  bar  perhaps  |  X  f  inches 
and  15  inches  long  is  bent  when  cold  either  by  pressure  or  by 
blows  of  a  hammer.  Bridge  iron  should  bend  through  an 
angle  of  90  degrees  to  a  curve  whose  radius  is  twice  the  thick- 
ness of  the  bar,  without  cracking.  Rivet  iron  should  bend 
through  1 80  degrees  until  the  sides  of  the  bar  are  in  contact, 
without  showing  signs  of  fracture.  Wrought  iron  that  breaks 
under  this  test  is  lacking  in  both  strength  and  ductility. 

The  tensile  test  on  a  small  specimen  not  less  than  12  inches 


ART.  88. 


WROUGHT  IRON. 


183 


in  length  and  0.5  square  inches  in  cross-section  is  mainly  em- 
ployed. The  elongation  should  be  measured  on  a  length  not 
less  than  8  inches.  The  following  requirements  for  structural 
iron  in  tension  are  those  recommended  by  a  committee  of  the 
American  Society  of  Civil  Engineers  in  1895  : 


Kind  of  Wrought  Iron. 

Yield  Point. 
Lbs.  per 
sq.  in. 

Maximum 
Strength. 
Lbs.  per 
sq.  in. 

Ultimate 
Elongation. 
P.er  cent. 

Reduction 
of  Area. 
Per  cent. 

Bars 

26000 

50000 

20 

30 

Tension  Plates  and  Shapes 

26000 

48000 

15 

20 

Compression  Plates  and  Shapes 

26000 

48  ooo 

12 

16 

Web  Plates 

26OOO 

46000 

8 

12 

The  term  'yield  point '  is  here  used  instead  of  elastic  limit,  as 
the  latter  is  strictly  the  point  at  which  the  elongation  ceases 
to  be  proportional  to  the  applied  load,  and  it  is  not  easy  to  de- 
termine its  exact  value.  On  a  delicate  machine,  however,  the 
sudden  increase  of  elongation  soon  after  the  elastic  limit  is 
passed  cannot  fail  to  be  noticed  by  the  drop  of  the  scale 
beam,  and  this  is  called  the  yield  point.  It  is  further  recom- 
mended that  the  stress  be  applied  to  the  specimen  at  the  uni- 
form rate  of  from  4  ooo  to  5  ooo  pounds  per  square  inch  per 
minute  below  the  yield  point  and  7  ooo  to  8  ooo  pounds  per 
square  inch  per  minute  above  the  yield  point. 

The  process  of  manufacture,  as  well  as  the  quality  of  the 
pig  iron,  influences  the  strength  of  wrought  iron.  The  higher 
the  percentage  of  carbon  the  greater  is  the  strength.  Best 
iron  is  10  per  cent  stronger  than  ordinary  merchant  iron  owing 
to  the  influence  of  the  second  rolling.  Cold  rolling  causes  a 
marked  increase  in  elastic  limit  and  ultimate  strength,  but  a 
decrease  in  ductility  or  ultimate  elongation.  Annealing  lowers 
the  ultimate  strength,  but  increases  the  elongation.  Boiler- 
plate iron  will  generally  have  an  ultimate  strength  greater  than 
55  ooo  pounds  per  square  inch.  Iron  wire,  owing  to  the  pro- 


184  THE   STRENGTH   OF  MATERIALS.  CH.  VIII. 

cess  of  drawing,  has  a  high  tensile  strength,  sometimes  greater 
than  100  ooo  pounds  per  square  inch. 

Good  wrought  iron  when  broken  by  tension  shows  a  fibrous 
structure.  If,  however,  it  be  subject  to  shocks  or  to  repeated 
stresses  which  exceed  the  elastic  limit,  the  molecular  structure 
becomes  changed  so  that  the  fracture  is  more  or  less  crystalline. 
The  effect  of  a  stress  slightly  exceeding  the  elastic  limit  is  to 
cause  a  small  permanent  set,  but  the  elastic  limit  will  be  found 
to  be  higher  than  before.  This  is  decidedly  injurious  to  the 
quality  of  the  material  on  account  of  the  accompanying  change 
in  structure,  and  hence  it  is  a  fundamental  principle  that  the 
working  unit-stresses  should  not  exceed  the  elastic  limit.  For 
proper  security  indeed  the  allowable  unit-stress  should  seldom 
be  greater  than  one  half  the  elastic  limit. 

In  a  rough  general  way  the  quality  of  wrought  iron  may  be 
estimated  by  the  product  of  its  tensile  strength  and  ultimate 
elongation,  this  product  being  an  approximate  measure  of  the 
work  required  to  produce  rupture.  A  more  precise  measure 
of  this  work  K  is,  however,  given  by  the  formula 

(15)  K=WS.  +  2$) 

in  which  s  is  the  ultimate  unit-elongation,  Se  is  the  elastic  limit, 
and  St  the  tensile  strength.  For  example,  take  two  wrought- 
iron  specimens,  the  first  having  Se  =  27  ooo,  5  =  66000,  s  =  13 
per  cent  =  0.13,  and  the  second  having  Se  =  24000,  St  =  51  OOO, 
s  =  24  per  cent  =  0.24.  Then  the  values  of  K  are 

K,  —  6  890,  K.,  =  10  080, 

which  shows  that  the  work  required  to  rupture  the  second 
specimen  will  be  much  greater  than  for  the  first.  Thus  high 
tensile  strength  is  not  a  good  quality  when  accompanied  by 
low  elongation.  The  value  of  K  is  the  number  of  inch-pounds 
of  work  required  for  the  rupture  of  a  specimen  one  square  inch 
in  section-  and  one  inch  long. 


ART.  89.  STEEL.  185 

Wrought  iron  has  no  proper  modulus  of  rupture  since,  ow- 
ing to  the  change  of  molecular  condition  after  the  elastic  limit 
is  exceeded,  bars  can  be  bent  to  almost  any  extent  without 
fracture.  The  same  is  true  of  soft  and  medium  steel. 

ART.  89.    STEEL. 

Steel  was  originally  produced  directly  from  pure  iron  ore  by 
the  action  of  a  hot  fire,  which  did  not  remove  the  carbon  to  a 
sufficient  extent  to  form  wrought  iron.  The  modern  processes, 
however,  involve  the  fusion  of  the  ore,  and  the  definition  of 
the  United  States  law  is  that  "  steel  is  iron  produced  by  fusion 
by  any  process,  and  which  is  malleable."  Chemically,  steel  is  a 
compound  of  iron  and  carbon  generally  intermediate  in  com- 
position between  cast  and  wrought  iron,  but  having  a  higher 
specific  gravity  than  either.  The  following  comparison  points 
out  the  distinctive  differences  between  the  three  kinds  of  iron: 

Per  cent  of  Carbon.  Spec.  Grav.  Properties. 

Cast  iron,  5       to  2  7.2      Fusible,  not  malleable. 

Steel,  i.SOtoo.io         7.8      Fusible  and  malleable. 

Wrought  iron,       0.30  to  0.05         7.7      Malleable,  not  fusible. 

It  should  be  observed  that  the  percentage  of  carbon  alone 
is  not  sufficient  to  distinguish  steel  from  wrought  iron  ;  also, 
that  the  mean  values  of  specific  gravity  stated  are  in  each  case 
subject  to  considerable  variation. 

The  three  principal  methods  of  manufacture  are  the  cruci- 
ble process,  the  open-hearth  process,  and  the  Bessemer  pro- 
cess. In  the  crucible  process  impure  wrought  iron  or  blister 
steel,  with  carbon  and  a  flux,  are  fused  in  a  sealed  vessel  to 
which  air  cannot  obtain  access ;  the  best  tool  steels  are  thus 
made.  In  the  open-hearth  process  pig  iron  is  melted  in  a 
Siemens  furnace,  wrought-iron  scrap  being  added  until  the 
proper  degree  of  carbonization  is  secured.  In  the  Bessemer 
process  pig  iron  is  completely  decarbonized  in  a  converter  by 


186  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

an  air  blast  and  then  recarbonized  to  the  proper  degree  by  the 
addition  of  spiegeleisen.  The  metal  from  the  open-hearth 
furnace  or  from  the  Bessemer  converter  is  cast  into  ingots, 
which  are  rolled  in  mills  to  the  required  forms.  The  open- 
hearth  process  produces  steel  for  guns,  armor  plates,  and  for 
some  structural  purposes ;  the  Bessemer  process  produces 
steel  for  railroad  rails  and  also  for  structural  shapes. 

The  physical  properties  of  steel  depend  both  upon  the 
method  of  manufacture  and  upon  the  chemical  composition, 
the  carbon  having  the  controlling  influence  upon  strength. 
Manganese  promotes  malleability  and  silicon  increases  the 
hardness,  while  phosphorus  and  sulphur  tend  to  cause  brittle- 
ness.  The  higher  the  percentage  of  carbon  within  reasonable 
limits  the  greater  is  the  ultimate  strength  and  the  less  the 
elongation.  THURSTON  proposes  the  formula 

(16)  St  =  60  ooo  -f-  70  oooC 

for  the  ultimate  tensile  strength  in  pounds  per  square  inch,  C 
being  the  per  cent  of  carbon.  Thus,  with  0.40  per  cent  of  car- 
bon the  value  of  St  is  88  ooo  pounds  per  square  inch.  The 
ultimate  elongation  is  approximately  inversely  proportional  to 
the  tensile  strength,  a  formula  frequently  given  being  : 

150           1500000 
elongation  in  per  cent  =  2-— r — -^  =  F ; 

thus  when  C  =  0.70,  the  tensile  strength  is  about  109  ooo 
pounds  per  square  inch  and  the  elongation  about  14  per  cent. 
These  approximate  formulas  refer  only  to  unannealed  steel. 

A  classification  of  steel  according  to  the  percentage  of 
carbon  and  its  physical  properties  of  tempering  and  welding  is 
as  follows : 

Extra  hard,  i.oo  to  o.6o#  C.,  takes  high  temper,  but  not  weldable. 

Hard,  0.70  to  0.40*  C.,  temperable,  but  welded  with  difficulty. 

Medium,       0.50  to  0.20*  C.,  poor  temper,  but  weldable. 

Mild,  0.40  to  0.05^  C.,  not  temperable,  but  easily  welded. 


ART.  89. 


STEEL. 


187 


It  is  seen  that  these  classes  overlap  so  that  there  are  no  distinct 
lines  of  demarcation.  The  extra-hard  steels  are  used  for  tools, 
the  hard  steels  for  piston  rods  and  other  parts  of  machines, 
the  medium  steels  for  rails  and  beams,  and  the  mild  or  soft 
steels  for  plates,  rivets,  and  other  purposes. 

The  influence  of  annealing,  or  keeping  the  metal  in  contact 
with  a  light  fire  for  some  days,  is  marked  in  reducing  the  ulti- 
mate strength  and  increasing  the  elongation.  As  an  example 
the  following  table  gives  some  of  the  results  of  a  large  series 
of  tests  exhibited  by  the  Bethlehem  Iron  Company  at  the 
World's  Columbian  Exposition  of  1893,  all  being  flat  bars  of 


Per  cent 
of 
Carbon. 

Tensile  Strength. 
Pounds  per  square  in. 

Ultimate  Elongation. 
Per  cent. 

Unan- 
nealed. 

Annealed. 

Unan- 
nealed  . 

Annealed. 

0.08 

58  ooo 

56000 

27 

31 

0.25 

&40OO 

75000 

21 

25 

0.50 

125  ooo 

9QOOO 

II 

19 

0.67 

136000 

112  OOO 

6 

16 

1.04 

153000 

I2800O 

3i 

II 

Bessemer  steel.  The  process  of  annealing  is  thus  seen  to 
greatly  improve  the  capacity  of  the  high-carbon  steels  to  resist 
work,  since  the  product  of  tensile  strength  and  elongation  is 
materially  increased  (Art.  88). 

For  bridges  and  buildings  the  following  requirements  are 
recommended  by  the  committee  report  of  the  American  So- 
ciety of  Civil  Engineers  in  1895,  for  ultimate  tensile  strength : 

For  low  steel,  60  ooo  ±  4  ooo, 

For  medium  steel,  65  ooo  ±  4  ooo, 

For  high  steel,  70000  ±  4000, 

all  being  in  pounds  per  square  inch.  The  yield  point  is  re- 
quired to  be  55  per  cent  of  the  ultimate  strength,  the  per  cent 
of  elongation  to  be  I  500000  divided  by  the  ultimate  strength, 


188  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

and  the  per  cent  of  reduction  of  area  to  be  2  800  ooo  divided 
by  the  ultimate  strength.  Specimens  of  medium  steel  cut 
from  bars  or  plates  must  stand  bending  through  180  degrees 
to  an  inner  radius  of  one  and  one  half  times  the  thickness  of 
the  specimen  without  sign  of  fracture,  while  for  the  high  steel 
the  same  must  be  the  case  to  a  radius  of  twice  the  thickness  of 
the  specimen. 

The  strength  of  steel  may  be  greatly  increased  by  com- 
pressing it  while  fluid,  by  the  use  of  nickel  as  an  alloy,  and  by 
the  processes  of  forging.  Nickel  steel  has  been  made  with  an 
elastic  limit  over  100000  and  with  an  ultimate  tensile  strength 
of  277  ooo  pounds  per  square  inch.  By  tempering  both 
strength  and  hardness  are  increased,  and  by  annealing  its  re- 
sistance to  shock  is  improved.  See  paper  by  R.  W.  DAVEN- 
PORT in  Engineering  News  for  Nov.  23,  1893. 

The  compressive  strength  of  steel  is  always  higher  than  the 
tensile  strength.  The  maximum  value  recorded  for  hardened 
steel  is  392000  pounds  per  square  inch.  The  expense  of  com- 
mercial tests  of  compression  is,  however,  so  great  that  they 
are  seldom  made.  The  shearing  strength  is  about  three-fourths 
of  the  tensile  strength.  Soft  and  structural  steels  have  no 
modulus  of  rupture,  since  bars  can  be  bent  through  180  degrees 
by  transverse  pressure. 

The  elastic  limit  is  usually  well  defined  and  closely  coinci- 
dent with  the  yield  point.  In  tension,  as  a  rough  rule,  it  may 
be  taken  as  one-half  of  the  ultimate  strength,  in  compression 
at  somewhat  less  than  one-half,  perhaps  one-third,  for  the  hard 
steels.  The  coefficient  of  elasticity  is  subject  to  but  little 
variation  with  the  percentage  of  carbon,  and  the  mean  value  of 
30000000  pounds  per  square  inch  may  be  used  in  ordinary 
computations  both  for  tensile  and  compressive  stresses  that  do 
not  exceed  the  elastic  limit.  In  shearing  the  coefficient  of  elas- 
ticity may  be  taken  as  a  mean  at  two-fifths  of  that  for  tension. 


ART.  90.  OTHER   MATERIALS.  189 

Steel  castings  are  extensively  used  for  axle  boxes,  cross- 
heads,  and  joints  in  structural  work.  They  contain  from  0.25 
to  0.50  per  cent  of  carbon,  ranging  in  tensile  strength  from 
60000  to  looooo  pounds  per  square  inch. 

Steel  has  entirely  supplanted  wrought  iron  for  railroad  rails, 
and  largely  so  for  structural  purposes.  Its  price  being  the 
same,  its  strength  greater,  its  structure  more  homogeneous,  the 
low  and  medium  varieties  are  coming  more  and  more  into  use 
as  a  satisfactory  and  reliable  material  for  large  classes  of  engi- 
neering constructions. 

ART.  90.    OTHER  MATERIALS. 

Concrete,  composed  of  hydraulic  mortar  and  broken  stone, 
is  an  ancient  material,  having  been  extensively  used  by  the 
Romans.  It  is  mainly  employed  for  foundations  and  mono- 
lithic structures,  but  in  some  cases  large  blocks  have  been 
made  which  are  laid  together  like  masonry.  Like  mortar,  its 
strength  increases  with  age.  When  six  months  old  its  mean 
compressive  strength  ranges  from  700  to  I  500  pounds  per 
square  inch,  and  when  one  year  old  it  is  probably  about  fifty 
per  cent  greater. 

Beton  is  an  artificial  stone  made  of  hydraulic  cement  and 
sand  which  has  been  subject  to  prolonged  trituration.  Its 
strength  is  about  double  that  of  ordinary  concrete. 

Ropes  are  made  of  hemp,  of  manilla,  and  of  iron  or  steel 
wire  with  a  hemp  center.  A  hemp  rope  one  inch  in  diameter 
has  an  ultimate  strength  of  about  6  ooo  pounds,  and  its  safe 
working  strength  is  about  800  pounds.  A  manilla  rope  is 
slightly  stronger.  Iron  and  steel  ropes  one  inch  in  diameter 
have  ultimate  strengths  of  about  36000  and  50000  pounds 
respectively,  the  safe  working  strengths  being  6  ooo  and  8  ooo 
pounds.  As  a  fair  rough  rule,  the  strength  of  ropes  may  be 
said  to  increase  as  the  squares  of  their  diameters. 


IQO  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

Phosphor  bronze  is  an  alloy  of  copper  and  tin  containing 
from  2  to  6  per  cent  of  phosphorus.  It  is  remarkable  for  its 
complete  fluidity  so  that  most  perfect  castings  caii  be  made. 
It  has  been  used  for  journal  bearings,  valve  seats,  and  even  for 
cannon.  It  is  hard  and  tough,  and  its  ultimate  tensile  strength 
may  range  from  40000  to  100000  pounds  per  square  inch. 

Aluminum  is  a  silver-gray  metal  which  is  malleable  and  duc- 
tile and  not  liable  to  corrode.  Its  specific  gravity  is  about 
2.65,  so  that  it  is  light,  weighing  only  168  pounds  per  cubic 
foot.  Its  ultimate  tensile  strength  is  about  25  ooo  pounds 
per  square  inch.  It  has  a  low  coefficient  of  elasticity,  and  its 
ultimate  elongation  is  also  low.  Alloys  of  aluminum  and 
copper  have  been  made  with  a  tensile  strength  and  elongation 
exceeding  those  of  wrought  iron,  but  have  not  come  into  use 
as  structural  materials. 

Numerous  brasses  and  bronzes  composed  of  copper,  tin, 
and  zinc  have  been  made.  The  strongest  was  ascertained  by 
THURSTON  to  be  that  composed  of  55  parts  of  copper,  43  of 
zinc,  and  2  of  tin,  its  ultimate  tensile  strength  of  68  900  pounds 
per  square  inch,  with  an  elongation  of  48  per  cent  and  a  reduc- 
tion of  area  of  70  per  cent.  See  THURSTON'S  Materials  of 
Engineering,  Vol.  III. 

Brass,  which  is  composed  of  copper  and  zinc,  is  almost  the 
only  alloy  which  has  come  into  extensive  use  in  the  arts  and 
•  which  at  the  same  time  is  a  fully  reliable  material.  In  the  form 
of  castings  it  has  a  tensile  strength  of  about  20  ooo  pounds  per 
square  inch,  in  the  form  of  rolled  sheets  or  wire  it  has  a  much 
greater  strength.  Brass  water-pipes  are  now  frequently  used 
in  houses  by  those  who  can  afford  to  pay  as  high  a  price  as 
20  cents  per  pound. 

The  strength  of  lead  is  only  about  one-tenth  of  that  of 
brass,  and  it  attains  a  permanent  set  under  a  small  tensile 

stress. 


ART.  91.  THE   FATIGUE   OF   MATERIALS  19! 

Glass  has  a  tensile  strength  of  about  5  ooo  and  a  compres- 
sive  strength  of  about  8  ooo  pounds  per  square  inch. 

ART.  91.    THE  FATIGUE  OF  MATERIALS. 

The  ultimate  strength  Su  is  usually  understood  to  be  that 
steady  unit-stress  which  causes  rupture  at  one  application. 
Experience  and  experiments,  however,  teach  that  if  a  unit- 
stress  somewhat  less  than  Su  be  applied  a  sufficient  number  of 
times  to  a  bar  rupture  will  be  caused.  The  experiments  of 
WOHLER  have  been  of  great  value  in  establishing  the  laws 
which  govern  the  rupture  of  metals  under  repeated  applica- 
tions of  stress.  For  instance,  he  found  that  the  rupture  of  a 
bar  of  wrought  iron  by  tension  was  caused  in  the  following 
different  ways. 

By  800  applications  of  52  800  pounds  per  square  inch. 

By        107  ooo  applications  of  48  400  pounds  per  square  inch. 

By       450  ooo  applications  of  39  ooo  pounds  per  square  inch. 

By  10  140  ooo  applications  of  35  ooo  pounds  per  square  inch. 

The  range  of  stress  in  each  of  these  applications  was  from  o  to 
the  designated  number  of  pounds  per  square  inch.  Here  it  is 
seen  that  the  breaking  stress  decreases  as  the  number  of  appli- 
cations increase.  In  other  experiments  where  the  initial  stress 
was  not  o,  but  a  permanent  value  5,  the  same  law  was  seen  to 
hold  good.  It  was  further  observed  that  a  bar  could  be  strained 
from  o  up  to  a  stress  near  its  elastic  limit  an  enormous  number 
of  times  without  rupture.  From  a  discussion  of  these  numer- 
ous experiments  the  following  laws  may  be  stated. 

1.  By  repeated  applications  of  stress  rupture  may  be  caused 
by  a  unit-stress  less  in  value  than  the  ultimate  strength 
of  the  material. 

2.  The  greater  the  range  of  stress  the  less  is  the  unit-stress 
required  to  produce  rupture  after  an  enormous  number 
of  applications. 

3.  When   the    unit-stress   in    a   bar   varies   from   O   up 


I92  THE   STRENGTH    OF   MATERIALS.  CH.  VIII. 

to  the  elastic  limit  the  number  of  applications  required 
to  rupture  it  is  enormous. 

4.  A  range  of  stress  from  tension  into  compression,  or  vice 
versa,  produces   rupture  with  a  less  number  of  applica- 
tions than  the  same  range  in  stress  of  one  kind  only. 

5.  When  the  range  of  stress  in  tension  is  equal  to  that  in 
compression  the  unit-stress  that  produces  rupture  after 
an  enormous  number  of  applications  is  a  little  greater 
than  one-half  the  elastic  limit. 

The  term  *  enormous  number '  used  in  stating  these  laws 
means  about  40  millions,  that  being  roughly  the  number  used 
by  WOHLER  to  cause  rupture  under  the  conditions  stated. 
For  all  practical  cases  of  repeated  stress,  except  in  fast  moving 
machinery,  this  great  number  would  seldom  be  exceeded  during 
the  natural  life  of  the  piece. 

In  Art.  8  it  was  recognized  that  the  working  stress  should 
be  less  for  pieces  subject  to  varying  stresses  than  for  those  car- 
rying steady  loads  only.  For  many  years  indeed  it  has  been 
the  practice  of  designers  to  grade  the  working  stress  accord- 
ing to  the  range  of  stresses  to  which  it  might  be  liable  to  be 
subjected.  WOHLER's  laws  and  experiments  afford  however  a 
means  of  grading  these  values  in  a  more  satisfactory  manner 
than  mere  judgment  can  do,  and  formulas  for  that  purpose  will 
be  deduced  in  the  next  Article.  After  the  working  stress 
Sw  is  determined  the  cross-section  of  the  piece  is  found  in  the 
usual  way,  if  in  tension  by  formula  (i),  and  if  in  compression 
by  formula  (i)  or  (10)  as  the  case  may  require. 

Prob.  136.  How  many  years  will  probably  be  required  for  a 
tie  bar  in  a  bridge  truss  to  receive  40  million  repetitions  of 
stress  ? 

ART.  92.    REPEATED  STRESSES. 

Consider  a  bar  in  which  the  unit-stress  varies  from  S'  to  S, 
the  latter  being  the  greater  numerically.  Both  S'  and  5  may 


ART.  92. 


REPEATED   STRESSES. 


193 


be  tension  or  both  may  be  compression,  or  one  may  be  ten- 
sion and  the  other  compression ;  in  the  last  case  the  sign  of 
6"  is  to  be  taken  as  minus.  Consider  the  stress  to  be  re- 
peated an  enormous  number  of  times  from  S'  to  S  and  rupture 
to  then  occur  under  the  greater  unit-stress  S.  By  the  sec- 
ond law  above  stated  5  is  some  function  of  5  —  S' ;  this  is 
equivalent  to  saying  that  5  is  a  function  of  5(i  —  S'/S),  or 
more  simply  a  function  of  S'/S.  Now  if  P'  and  P  be  the 
total  stresses  on  the  bar  the  ratio  S'/S  equals  P'/P,  and 
hence  the  unit-stress  5  which  causes  rupture  after  an  enor- 
mous number  of  repetitions  is  a  function  of  P'/P. 

LAUNHARDT'S  formula  for  5  applies  to  the  case  where  the 
limiting  stresses  Pr  and  P  are  both  tension  or  both  compres- 
sion, so  that  P'/P  is  always  positive.  Let  the  values  of  this 
ratio  be  taken  as  abscissas  ranging  from  o  to  I,  and  those  of 
S  as  ordinates.  Let  the  function  of  P'/P  be  supposed  to  rep- 
resent a  straight  line  whose  equation  is 

5  =  m  -f-  n-p- , 

in  which  m  and  n  are  constants  to  be  determined.     Let  u  be 

the  ultimate  strength  of  the  material  under  one  application  of 

stress,  and  e  the  unit-stress 

at  the  elastic  limit.      Now 

if  P'/P  is  unity,  then  5  is 

u    and    hence    u  =  m  -f-  n. 

Also,  from  the  third  law  of 

the  last  article,    if  P'/P  is 

zero,  then  5  is  e  and  hence 

e  =  m.       Accordingly    the 

value  of  m  is  e,  that  of  n  is  FlG-  56. 

u  —  e>  and  the  equation  of  the  straight  line  becomes 


—1.0 


-0.5 


-H).5 


+1.0 


which  gives  the  unit-stress  5  that  ruptures  the  bar  after  an 


194  THE   STRENGTH   OF   MATERIALS.  Cli    VIII. 

enormous  number  of  repetitions  of  stress  ranging  from  P'  to  P. 
For  mean  values  of  u  and  ^,  this  becomes  for  wrought  iron 

S=  25000(1  +ff 

For  example,  let  a  bar  range  in  tension  from  80000  to 
160000  pounds;  then  the  value  of  P'/P  is  %  and,  from  the 
formula,  S  =  40  ooo  pounds  per  square  inch  is  the  rupturing 
unit-stress. 

WEYRAUCH'S  formula  for  5  applies  to  the  case  where  the 
bar  ranges  in  stress  from  P'  to  P,  one  being  tension  and  the 
other  compression,  and  P  being  the  greater  numerically. 
Here  P'/P  is  always  negative,  and  the  law  connecting  it  with 
»S  is  taken  to  be  the  same  as  before.  Let  e  be  the  unit-stress 
at  the  elastic  limit  and  /the  unit-stress  which,  under  the  fifth 
law  of  the  last  article,  causes  rupture  when  P'  and  P  are  nu- 
merically equal.  By  the  third  law,  if  P'/P  is  zero,  then  5  is 
e  and  hence  e  =  m.  By  the  fifth  law,  if  P'/P  is  —  I,  then  5 
is  /and  hence  /=  m  —  n.  'Accordingly  the  value  of  m  is  e, 
that  of  n  is  e  —/,  and  the  equation  of  the  straight  line  is 


For  example,  taking  /  =  %e,  this  becomes  for  wrought  iron 


in  which  the  value  of  P'/P  is  to  be  inserted  as  negative. 
Thus  if  P1  =  80  ooo  pounds  compression  and  P=  160000 
pounds  tension,  then  P'/P=  —  £,  and  5=18750  pounds 
per  square  inch  is  the  rupturing  unit-stress  after  an  enormous 
number  of  repetitions. 

In  Fig.  56  the  ordinates  u,  e,  and  /  represent  the  values  of 
5  for  the  values  +  I,  o,  and  —  I  of  the  ratio  P'/P.  The 
straight  line  joining  the  tops  of  the  ordinates  u  and  e  repre- 
sents LAUNHARDT'S  formula,  while  that  joining  the  tops  of 
the  ordinates  e  and  f  represents  WEYRAUCH'S  formula. 


ART.  92.  REPEATED   STRESSES.  IQ5 

Another  formula  may  be  established  by  assuming  the  law 
of  variation  of  5  to  be  that  represented  by  a  curve  joining  the 
tops  of  the  three  ordinates  u,  e,  and  f.  The  simplest  curve 
is  a  parabola  whose  equation  is 

P'  P'\* 


To  determine  ;//,  n,  and/,  consider  first  that  if  P'/P=  -f-  i, 
then  S=u  and  hence  u—m-\-n-^-p\  second  that  if 
P'/P=o,  then  S=e  and  hence  e  =  m\  third  that  if 
P'/P  —  —  i,  then  5=/and  hence/=  m  —  n+p.  From 
these  three  conditions  the  values  of  m,  n,  and  p  are  found 
and  accordingly 

—2e 


is  a  formula  for  the  rupturing  unit-stress  in  a  bar  whose  total 
stress  ranges  between  the  limits  P'  and  P.  If  P'  and  P  are 
both  tension  or  both  compression  the  ratio  P'/P  is  positive, 
if  one  is  tension  and  the  other  compression  the  ratio  P'/P  is 
negative.  It  is  seen  that  (17)  always  gives  values  of  5  a  little 
smaller  than  those  found  from  the  straight-line  formulas. 

For  structural  steel  where  u  =  64000,  e  =  32  ooo,  and/  = 
16000  pounds  per  square  inch,  the  formula  (17)  becomes 


For  a  bar  of  such  steel  whose  stress  ranges  from  1  80000 
pounds  tension  under  dead  load  to  540000  pounds  tension 
under  live  load  the  value  of  P'/P  is  +  £,  and  the  formula 
gives  5  =  40900  pounds  per  square  inch.  If  the  stress  ranges 
from  180000  pounds  compression  to  540000  pounds  tension, 
then  the  value  of  P'/P  is  —  £,  and  the  formula  gives  5  = 
24900  pounds  per  square  inch. 

In  using  these  formulas  in  cases  of  design  a  factor  of  safety 


196  THE   STRENGTH   OF   MATERIALS.  CH.  VIII. 

is  applied.     Thus  with  a  factor  of  4  the  straight-line  formulas 
for  structural  steel  take  the  form 


=  8000(1+     -, 


the  first  being  used  when  P'/P  is  positive  and  the  second 
when  it  is  negative,  while  the  parabolic  formula  reduces  to 


These  formulas  give  the  allowable  unit-stresses  to  be  used  in 
designing  bridge  members  of  structural  steel.  As  an  example 
let  it  be  required  to  find  by  help  of  the  last  formula  the  proper 
sectional  area  for  a  bar  of  structural  steel  subject  to  repeated 
tension  ranging  between  29000  and  145000  pounds;  here 
P'  /P  •=.  -f-  O.2,  5  =  9280  pounds  per  square  inch,  and 

P        145000 
A=S'~      9280       =  IS      scluare  mches. 

In  like  manner  if  a  short  steel  bar  ranges  in  stress  from 
29000  pounds  compression  to  145000  pounds  tension  the 
value  of  P'/P  is  —  0.2  and  S  =  6  880  pounds  per  square  inch  ; 
then  the  cross-section  required  is  21.1  square  inches. 

Prob.  137.  A  short  bar  of  wrought  iron  is  subject  to  re- 
peated stresses  ranging  from  16000  pounds  compression  to 
80  ooo  pounds  tension.  What  should  be  the  area  of  its  cross- 
section  for  a  factor  of  safety  of  5  ? 

Prob.  138.  A  wrought-iron  bar  has  a  tension  of  3000 
pounds  per  square  inch  as  its  least  unit-stress.  For  a  factor 
of  safety  of  4  what  is  the  greatest  tensile  unit-stress  5  to 
which  it  should  be  subjected  when  the  unit-stress  is  often 
repeated  from  3  ooo  up  to  5  ? 


ART.  93.  SUDDEN   LOADS  AND   IMPACT.  197 


CHAPTER   IX. 
THE   RESILIENCE   OF   MATERIALS. 

ART.  93.  SUDDEN  LOADS  AND  IMPACT. 
When  a  tensile  load  is  slowly  and  uniformly  applied  to  a  bar  it 
increases  slowly  from  o  up  to  the  final  value  P,  and  the  stress 
in  the  bar  at  any  instant  is  equal  to  the  tensile  force  existing 
at  that  instant;  the  elongation  of  the  bar  increases  propor- 
tionally to  the  stress  from  o  up  to  the  final  limit  A,  if  the 
elastic  limit  is  not  exceeded.  The  work  done  upon  the  bar  by 
the  external  force  is  then  equal  to  its  mean  intensity  %P  multi- 
plied by  the  distance  A,  or  \P\ ;  the  work  of  the  molecular 
forces  is  also  equal  to  this  same  quantity  £PA. 

A  load  P  is  said  to  be  suddenly  applied  when  its  intensity  is 
the  same  from  the  beginning  to  the  end  of  the  elongation. 
The  stress  in  the  bar,  however,  increases  from  o  up  to  a  limit  Q. 
Let  y  be  the  elongation  produced  by  the  sudden  load  P\  then 
the  work  of  this  external  force  is  Py.  If  the  stresses  are 
within  the  elastic  limit  so  that  they  increase  proportionally  to 
the  elongation,  the  mean  stress  is  \Q  and  the  work  of  the  re- 
sisting forces  is  \Qy.  Hence,  as  these  two  works  must  be  equal, 

\Qy  =  Py         or         Q  =  2P. 

Now  let  A  be  the  elongation  due  to  the  load  P  when  gradually 
applied,  then  by  law  (B), 

i=s    or        A 

Therefore  is  established  the  following  important  theoretical  law, 

A  suddenly  applied  load  produces  double  the  stress  and 
double  the  deformation  caused  by  the  same  load  when 
applied  slowly  with  uniform  increments. 


198  THE    RESILIENCE   OF   MATERIALS.  CH.  IX. 

This  law  is  only  true  when  all  the  stresses  are  within  the  elastic 
limit  of  the  material.  The  sudden  load  P  thus  causes  the  end 
of  the  bar  to  move  from  o  to  2A,  when  the  stress  becomes  2P 
the  resultant  force  tending  to  move  the  end  is  P  —  2P  or  —  P 
and  hence  the  end  moves  backward,  until  after  a  series  of 
oscillations  it  comes  to  rest  with  the  elongation  A  due  to  the 
static  stress  P.  The  time  of  this  oscillation,  as  also  the  velocity 
of  the  end  of  the  bar  at  any  instant,  can  be  computed  by  the 
principles  of  dynamics. 

Impact  is  said  to  be  produced  upon  the  end  of  a  bar  when 
a  load  P  falls  from  a  height  h  upon  it.  Here  the  stress  in  the 
bar  will  increase  from  o  up  to  a  certain  limit  Q  and  the  defor- 
mation from  o  up  to  a  certain  limit  y.  If  the  elastic  limit  of 
the  material  be  not  exceeded,  the  stress  at  any  instant  will  be 
proportional  to  the  deformation,  so  that  the  work  of  the  in- 
ternal stresses  will  be  %Qy.  The  work  done  by  the  exterior  force 
P  in  the  same  time  is  P(h  -\-  y).  Hence 


But  if  A  be  the  deformation  due  to  a  static  load  P,  the  law  of 
proportionality  gives 

Q    y 

p  ~A* 

Combining  these  two  equations  there  is  found, 


If  h  =  o  these  formulas  reduce  to  Q  =  2P  and  y  =  2A,  which 
is  the  case  of  a  suddenly  applied  load ;  if  h  =  4A,  they  become 
Q  =  4P  and  y  =  4/1 ;  if  h  =  I2A  they  give  Q  =  6P  and  y  =  6A. 
Since  A  is  a  small  quantity  for  any  metallic  bar,  it  follows  that 
a  load  P  dropping  from  a  moderate  height  may  produce  great 


ART.  94.  THE    MODULUS   OF   RESILIENCE. 

stresses  and  deformations.  Experiments  made  upon  springs 
show  that  the  theory  here  presented  is  correct,  provided  the 
elastic  limit  of  the  material  is  not  surpassed  by  the  stress  Q. 

The  effect  of  loads  applied  with  impact  is  therefore  to  cause 
stresses  and  deformations  greatly  exceeding  those  produced  by 
the  same  static  loads,  so  that  the  elastic  limit  may  perhaps  be 
often  exceeded.  Moreover  the  rapid  oscillations  and  the  rapid 
variations  in  the  stresses  cause  a  change  in  molecular  structure 
which  impairs  the  elasticity  of  the  material.  Generally  it  will 
be  found  that  the  appearance  of  a  fracture  of  a  bar  which  has 
been  subject  to  shocks  is  of  a  crystalline  nature,  whereas  the 
same  material,  if  ruptured  under  a  gradually  increasing  stress, 
would  exhibit  a  tough  fibrous  structure.  Shocks  which  produce 
stresses  above  the  elastic  limit  cause  the  material  to  become 
stiff  and  brittle,  and  hence  it  is  that  the  working  unit-stresses 
based  upon  static  loads  should  be  taken  very  low  (Art.  8). 

Prob.  139.  In  an  experiment  upon  a  spring  a  weight  of  14.79 
ounces  produced  an  elongation  of  0.42  inches,  but  when 
dropped  from  a  height  of  7.72  inches  it  produced  a  stress  of  102.3 
ounces  and  an  elongation  of  2.90  inches.  Compare  theory  with 
experiment.  - 

ART.  94.    THE  MODULUS  OF  RESILIENCE. 

When  an  applied  stress  causes  a  deformation  work  is  done. 
Thus  if  a  tensile  stress  P  be  applied  by  increments  to  a  bar, 
so  that  the  stress  gradually  increases  from  o  to  the  value  P,  the 
work  done  is  the  product  of  the  average  stress  by  the  total 
elongation  A.  This  product  is  termed  the  resilience  of  the  bar. 
If  the  stress  does  not  exceed  the  elastic  limit  of  the  material 
the  average  stress  is  \Py  and  the  work  or  resilience  is  \P\.  If 
the  cross-section  of  the  bar  be  A  and  its  length  /,  the  unit- 
stress  is  P -±  A  =  S  and  the  unit-elongation  is  X  -±-  /  =  j,  So 
that  the  work  of  the  internal  resisting  stresses  performed 


200  THE  RESILIENCE  OF   MATERIALS.  CH.  IX. 

on  each  unit  of  length  of  the  bar  per  unit  of  cross-section  is 
From  formula  (2)  the  value  of  s  is  -,  and  accordingly  this 


work  may  be  written, 

(.8)  K=1-^. 

If  5  be  the  unit-stress  at  the  elastic  limit,  the  quantity  K  is 
called  the  modulus  of  resilience  of  the  material. 

Resilience  is  often  regarded  as  a  measure  of  the  capacity  of 
a  material  to  withstand  impact,  for  if  a  shock  or  sudden  blow 
be  produced  by  a  falling  body,  its  intensity  depends  upon  the 
weight  and  the  height  through  which  it  has  fallen,  that  is,  upon 
its  kinetic  energy  or  work.  Hence  the  higher  the  resilience 
of  a  material  the  greater  is  its  capacity  to  endure  work  that 
may  be  performed  upon  it.  The  modulus  of  resilience  is  a 
measure  of  this  capacity  within  the  elastic  limit  only. 

The  following  are  values  of  the  modulus  of  resilience  as 
computed  from  (18)  by  the  use  of  the  average  constants  given 
in  Art.  5. 

For  timber,  K=    3.0  inch-pounds, 

For  cast  iron,  K=    1.2  inch-pounds, 

For  wrought  iron,        K  =  12.5  inch-pounds, 

For  steel,  K  =  41.7  inch-pounds. 

The  ultimate  resilience  of  materials  cannot  be  expressed  by  a 

rational  formula,  because  the  law  of  increase  of  elongation  be- 

yond the  elastic   limit  is  unknown.     In  Fig.   I    the    ultimate 

resilience  is  indicated  by  the  area  between  any  curve  and  the 

axis  of  abscissas,  since  that  area  has  the  same  value  as  the  total 

work  performed  in  producing  rupture.     For  timber  and  cast 

iron  the  ratio  of  these  areas  is  about  the  same  as  that  of  the 

values  of  K,  but  for  wrought  iron  and  steel  the  areas  are  nearly 

equal. 

Prob.  140.  What  horse-power  engine  is  required  to  strain  250 


ART.  95.  EXTERNAL   WORK   AND    RESILIENCE.  2OI 

times  per  minute  a  bar  of  wrought  iron  18  feet  long  and  2 
inches  in  diameter  from  o  up  to  12  500  pounds  per  square  inch. 

ART.  95.    EXTERNAL  WORK  AND  RESILIENCE. 

When  a  body  is  deformed  by  applied  forces  the  work  done 
by  these  forces  is  called  the  external  work.  For  example,  if 
a  bar  is  subjected  to  a  tensile  force  which  is  slowly  applied 
until  it  reaches  the  intensity  P,  an  elongation  A  is  produced, 
and  the  external  work  is  %PX.  Again,  if  a  beam  be  subject  to 
a  concentrated  load  P  gradually  applied,  a  deflection  A  occurs 
under  the  load  and  the  external  work  is  $PA.  If  the  load  is 
applied  suddenly  so  that  its  full  intensity  is  P  during  the  entire 
time  of  application,  then  the  external  works  in  the  two  cases 
are  P\  and  PA  respectively.  If  P  falls  from  a  height  //  above 
the  top  of  the  bar  or  beam  the  external  works  are  P(Ji  +  X) 
and  P(h  +  ^)  respectively. 

If  a  beam  be  uniformly  loaded  with  w  per  linear  unit  the 
load  on  any  short  length  dx  is  wdxy  and  if  y  be  the  deflection 
at  the  point  whose  abscissa  is  x,  the  elementary  external  work 
for  a  gradually  applied  load  is  \wy .  dx.  The  integration  of 
this  over  the  entire  length  of  the  beam  will  give  the  total  ex- 
ternal work  of  the  uniform  load. 

As  external  force  is  resisted  by  internal  stress  so  external 
work  is  resisted  by  internal  work.  Each  elementary  stress 
multiplied  by  its  displacement  gives  a  corresponding  elemen- 
tary work,  and  the  sum  of  all  these  products  is  the  total  inter- 
nal work.  By  the  law  of  conservation  of  energy, 

Internal  Work  =  External  Work, 

provided  that  no  work  is  lost  in  heat  by  the  application  of  the 
external  forces.  Now  the  word  *  resilience '  is  used  to  denote 
the  internal  work  of  the  stresses,  and  hence 

Resilience  =  External  Work, 


202  THE  RESILIENCE   OF  MATERIALS.  CH.  IX. 

or  the  resilience  of  a  body  is  its  capacity  to  resist  the  work  of 
external  forces. 

Elastic  resilience  is  internal  work  when  the  body  is  not 
stressed  beyond  the  elastic  limit.  Non-elastic  resilience  is  in- 
ternal work  when  the  stresses  range  from  the  elastic  limit  to 
the  point  of  rupture. 

In  Art.  94  an  expression  for  resilience  within  the  elastic 
limit  was  deduced  for  a  bar  of  unit  length  and  unit  cross-sec- 
tion. If  a  bar  of  cross-section  A  and  length  /  be  subject  to  a 
tensile  or  compressive  force  P,  the  deformation  A  is  produced. 
If  the  load  be  gradually  applied  the  external  work  is  ^P\. 
Let  S  be  the  unit-stress  produced,  and  E  be  the  coefficient  of 
elasticity  of  the  material.  Then,  from  Arts.  2  and  4, 

P=AS,  K=^; 

and  accordingly  the  internal  work  or  elastic  resilience  is 

(18)'  K=~.Al; 

that  is,  the  elastic  resiliences  of  bars  of  the  same  material  un- 
der the  same  unit-stress  are  proportional  to  their  volumes.  If 
6"  be  the  stress  at  the  elastic  limit  the  quantity  S1/2E  is  the 
modulus  of  resilience,  and  accordingly  the  maximum  elastic 
resilience  of  a  bar  is  the  product  of  its  modulus  of  resilience  by 
its  volume. 

A  theoretic  expression  for  non-elastic  resilience  cannot  be 
deduced,  but  in  Art.  97  it  will  be  shown  how  this  can  be  esti- 
mated when  sufficient  experimental  data  are  given.  Non- 
elastic  resiliences,  however,  are  generally  closely  proportional 
to  the  volumes  of  bodies,  the  material  and  the  maximum  stress 
being  constant. 

Prob.  141.  How  many  foot-pounds  of  work  are  required  to 
strain  a  wrought-iron  bar,  4  inches  in  diameter  and  54  inches 
long,  from  6000  pounds  per  square  inch  up  to  12000  pounds 
per  square  inch  ? 


ART.  96.  ELASTIC   RESILIENCE  OF  BEAMS.  2O3 

ART.  96.    ELASTIC  RESILIENCE  OF  BEAMS. 

When  a  beam  deflects  under  the  action  of  a  load  the  fibers 
on  one  side  of  the  neutral  surface  are  elongated  while  those 
on  the  other  side  are  shortened.  If  the  elastic  limit  is  not  ex- 
ceeded the  stress  in  any  fiber  is  proportional  to  its  distance 
from  the  neutral  surface  (Art.  20).  The  internal  work  or 
elastic  resilience  of  the  beam  is  the  half-sum  of  the  products 
formed  by  multiplying  the  stress  upon  each  elementary  area 
by  its  corresponding  change  of  length.  The  half-sum  instead 
of  the  sum  is  taken,  because  the  stress  uniformly  increases 
from  o  up  to  its  maximum  value  as  the  load  is  applied.  Thus, 
if  T  be  the  unit-stress  under  the  elongation  e,  the  unit-stress 

Tx 

for  an  elongation  x  is  — ,  and  the  work  in  the  distance  dx  is 

—dx;  integrating  this  between  the  limits  o  and  e  gives  \Te  as 

the  internal  work. 

Using  the  same  notation  as  in  Chapter  III,  the  horizontal 
unit-stress  upon  the  remotest  fiber  at  the  dangerous  section  of 
the  beam  is  called  S,  and  the  distance  of  that  fiber  from  the 
neutral  surface  is  called  c.  Let  a  single  concentrated  load  W 
be  gradually  applied  to  the  beam,  and  let  A  be  the  deflection 
beneath  it.  The  external  work  of  the  load  is  then  |  WA%  and 
this  equals  the  elastic  resilience  if  the  unit-stress  5  does  not 
surpass  the  elastic  limit.  If  /  be  the  length  of  the  beam,  and 
7  the  moment  of  inertia  of  the  cross-section,  the  value  of  W 
is,  from  Art.  29, 

*-•% 

where  n  is  I  for  a  cantilever  loaded  at  the  end  and  4  for  a 
simple  beam  loaded  at  the  middle.     Also  from  Art.  37, 

nsr 

A  = pr, 

mcE 


204  THE  RESILIENCE  OF  MATERIALS.  CH.  IX. 

where  m  is  3  for  the  cantilever  and  48  for  the  simple  beam. 
The  internal  work  of  the  beam  hence  is  : 


. 

2  E    me* 

or,  putting  for  /  its  value  Ar*  where  A  is  the  area  and  r  the 
least  radius  of  gyration  of  the  cross-section, 

o*  *-=•«•?•" 

which  is  a  general  expression  for  the  elastic  resilience  of  a 
beam  under  a  single  concentrated  load. 

1  5* 

If  the  beam  be  strained  to  the  elastic  limit  the  factor  -  -7, 

2  E 

is  the  modulus  of  resilience  of  the  material  (Art.  94).     For 

a 

either  the  cantilever  or  the  simple  beam  the  value  of  —  is  J. 

m 

r*  . 
For  a  rectangular  beam  —  a  is  J.     Thus  for  a  rectangular  beam 

the  internal  work  is 

(19)'  K  =  l2^Al=L*.Al, 

that  is,  the  work  required  to  deflect  a  rectangular  beam  by  a 
concentrated  load  is  proportional  to  its  volume  Aly  and  the 
work  required  to  cause  the  stress  5  to  reach  the  elastic  limit  is 
the  product  of  the  modulus  of  resilience  and  one-ninth  of  its 
volume. 

For  a  cantilever  loaded  uniformly  with  w  pounds  per  linear 
foot  the  load  on  any  short  length  dx  is  wdx,  and  if  y  be  the 
deflection  at  that  point  the  elementary  external  work  is  \wydx. 
Inserting  for/  its  value  from  Art.  34,  there  is  found 


24o£7        8afi/ 
for  the  total  external  work  of  the  uniform  load.     This  must  be 


ART.  96.  ELASTIC  RESILIENCE  OF  BEAMS.  2O5 

equal  to  the  internal  work.  Substituting  for  W  its  value  in 
terms  of  5  from  Art.  29,  the  elastic  resilience  of  a  rectangular 
cantilever  is  then 


which  is  nine-tenths  of  that  found  for  the  concentrated  load, 
and  also  proportional  to  the  volume  and  modulus  of  resilience. 
A  similar  result  is  easily  deduced  for  a  simple  beam  uniformly 
loaded. 

Formula  (19)  shows  that  the  internal  work  or  resilience  de- 
veloped within  the  elastic  limit  is  proportional  to  the  product 
of  the  volume  of  the  beam  and  the  ratio  r'/c*.  As,  however, 
this  ratio  always  has  a  numerical  value  which  is  the  same  for 
similar  sections,  it  may  be  stated  as  a  general  law,  that  the 
elastic  resiliences  of  beams  of  similar  cross-section  are  propor- 
tional to  their  volumes. 

As  a  numerical  example  let  it  be  required  to  determine  the 
horse-power  necessary  to  deflect  50  times  per  second  a  rectan- 
gular wrought-iron  beam  6  feet  long,  2  inches  wide,  and  3 
inches  deep,  so  that  at  each  deflection  the  unit-stress  5  may 
range  from  5000  to  10000  pounds  per  square  inch,  the  beam 
being  a  cantilever  with  the  load  applied  at  the  end.  From 
formula  (19)'  the  work  in  fifty  deflections  is 

K  =  S<X2X       (ioooo«  -  5  ooo«)  =  3  600  inch-pounds, 


which  is  300  foot-pounds,  and  hence  the  power  required  is 
300/550  =  0.52  horse-powers. 

The  strength  of  a  rectangular  beam  increases  with  the  square 
of  its  depth  and  its  stiffness  with  the  cube  of  the  depth  (Arts. 
29  and  36).  The  elastic  resilience,  however,  increases  only 
with  the  area  of  the  cross-section  ;  hence  for  a  given  unit-stress 
S  it  is  immaterial  whether  the  short  or  the  long  side  of  a  beam 


206 


THE   RESILIENCE  OF   MATERIALS. 


CH.  IX. 


be  placed  vertical  when  its  office  is  the  resistance  of  external 
work  only. 

Prob.  142.  Deduce  an  expression  for  the  elastic  resilience 
K  of  a  beam  fixed  at  both  ends  and  loaded  in  the  middle ;  also 
for  a  beam  fixed  at  both  ends  and  uniformly  loaded. 

ART.  97.    ULTIMATE  RESILIENCE. 

The  ultimate  resilience  of  a  body  under  stress  is  equal  to 
the  total  external  work  required  to  produce  rupture.  The 
elastic  resilience  is  that  part  of  the  ultimate  resilience  in  which 
the  stresses  do  not  surpass  the  elastic  limit  of  the  material. 
In  wrought  iron  and  steel  the  ultimate  resilience  greatly  sur- 
passes the  elastic  resilience,  being  sometimes  five  hundred 
times  as  large. 

In  order  to  show  this  fact  the  particular  case  of  a  steel  speci- 
men 12  inches  long  and  0.505  inches  in  diameter  will  be  taken, 
which  was  tested  in  a  tension  machine  and  the  elongations 
observed  at  certain  intervals.  In  the  following  table  the  first 


Load. 
Pounds. 

Stress. 
Pounds  per 
square  in. 

Elongation. 
Per  cent. 

Partial 
Work. 
Inch-Lbs. 

Total  Work 
Inch-Lbs. 

2OO 

IOOO 

0.00 

O 

IOOO 

5000 

O.OI 

0.3 

0 

3OOO 

I5OOO 

O.04 

3-0 

3 

6.0 

$000 

2500O 

O.O7 

9 

7000 

35000 

O.IO 

9.0 

18 

16.0 

9000 

45000 

O.I4 

34 

9600 

48000 

0.16 

9-3 

44 

10  000 

5OOOO 

0.7O 

264.6 

308 

12  000 

60000 

I.  QO 

660.0 

968 

I4OOO 

7OOOO 

3.62 

II20.O 

2088 

I6OOO 

80000 

8.50 

3660.0 

5748 

16800 

83600 

15-20 

5480.6 

11229 

I5OOO 

75000 

24.50 

7374  9 

18604 

ART.  97.  ULTIMATE   RESILIENCE.  2O/ 

column  gives  the  total  applied  load,  the  second  the  correspond- 
ing stress  per  square  inch,  and  the  third  the  elongation  ex- 
pressed in  per  cent.  The  elastic  limit  was  observed  at  48  ooo 
pounds  per  square  inch  with  o.io  per  cent  elongation.  The 
elongation  then  rapidly  increased  with  the  unit-stress  (as  seen 
in  the  diagram  Fig.  I  of  Art.  5).  At  83  600  pounds  per  square 
inch  the  maximum  tensile  strength  was  reached  and  the  mate- 
rial was  elongating  very  rapidly.  The  load  was  then  slowly 
removed,  but  the  elongation  continued  to  increase  until  rup- 
ture occurred  at  75  ooo  pounds  per  square  inch  with  a  total 
elongation  of  24.5  per  cent. 

The  external  work  per  cubic  inch  of  material  may  be  approxi- 
mately computed  for  any  interval  by  multiplying  the  average 
stress  during  that  interval  by  the  elongation  which  occurs. 
The  given  elongations  divided  by  100  give  the  elongations  per 
linear  inch.  Thus  while  the  load  ranged  from  i  ooo  to  3  ooo 
pounds  the  elongation  per  inch  increased  from  o.oooi  to  0.0004. 
Hence  the  work  done  upon  one  cubic  inch  of  the  specimen  in 
that  interval  was, 

£(15000  +  5  ooo)  (0.0004  —  0.0001)  =  3.0  men-pounds. 
Similarly  the  external  work  per  cubic  inch  performed  in  the 
last  interval  is 

$(83  600  +  75  ooo)  (0.245  —  0.152)  =  7374-9  inch  pounds, 
and  thus  the  quantities  in  the  fourth  column  of  the  table  arc 
separately  computed. 

The  summation  of  the  fourth  column  gives  the  total  exter- 
nal work  per  cubic  inch  required  to  stress  the  bar  from  o  up  to 
the  point  of  rupture.  In  the  last  column  sums  are  given  for 
each  value  of  the  unit-stress,  and  these  are  closely  equal  to  the 
internal  works.  Thus  it  is  seen  in  this  particular  case  that  for 
one  cubic  inch  of  material 

Elastic  Resilience      =  44  inch-pounds, 
Ultimate  Resilience  =  18604  inch-pounds; 


208  THE   RESILIENCE   OF   MATERIALS.  CH.  IX. 

and  therefore  that  the  ultimate  resilience  is  more  than  400 
times  as  great  as  the  elastic  resilience. 

The  total  external  work  required  to  rupture  the  specimen  is 
found  by  multiplying  the  above  ultimate  resilience  by  the  vol- 
ume of  the  specimen,  which  is  2.4  cubic  inches.  This  work  is 
found  to  be  3721  foot-pounds. 

The  example  here  given  is  a  case  of  static  resilience,  where 
the  internal  work  is  slowly  developed  under  the  action  of  an 
external  force  gradually  applied.  The  more  common  cases  of 
resilience,  however,  are  those  developed  by  the  impact  of  a 
falling  body,  and  a  general  discussion  of  these  will  be  presented 
in  the  following  articles. 

Prob.  143.  Compute  the  elastic  and  ultimate  static  resilience 
for  the  specimen  of  wrought  iron  whose  test  is  given  in  Art.  5, 
page  ii. 

ART.  98.    EARLY  HISTORY  OF  RESILIENCE. 

The  matter  in  the  remainder  of  this  chapter  is  taken  from 
an  address  on  *  The  Resistance  of  Materials  under  Impact/ 
delivered  by  the  author  in  August,  1894,  before  the  Section  of 
Mechanical  Science  and  Engineering  of  the  American  Associ- 
ation for  the  Advancement  of  Science. 

In  1807  THOMAS  YOUNG  announced  the  fundamental  ideas  of 
the  resistance  of  materials  under  impact.  "  The  action  which 
resists  pressure,"  he  said,  "  is  called  strength,  and  that  which 
resists  impulse  may  properly  be  called  resilience."  He  stated 
that  the  resilience  of  a  body  is  proportional  to  its  strength  and 
extension  jointly,  and  that  it  is  measured  by  the  height  through 
which  a  given  weight  must  fall  to  cause  rupture.  The  resili- 
ence of  beams  of  the  same  kind  he  made  proportional  to  their 
volumes,  as  also  the  resilience  of  shafts,  whether  solid  or  hol- 
low. He  further  suggested  that  a  very  high  velocity  of  the 


ART.  98.  EARLY   HISTORY   OF  RESILIENCE.  2OO, 

moving  weight  may  rupture  a  body  by  impact  before  its  full 
resilience  can  be  developed. 

Resilience  is  thus  the  capacity  of  a  body  to  resist  applied 
work,  or  it  is  the  internal  work  which  can  be  developed  by  the 
energy  of  a  moving  body.  External  force  is  resisted  by  inter- 
nal stress,  and  the  resulting  deformation  is  a  secondary  conse- 
quence ;  but  impact  is  resisted  by  resilience,  where  the  deforma- 
tion is  of  equal  importance  with  the  stress,  for  internal  work 
is  the  product  of  these  two  factors. 

In  YOUNG'S  time  the  elastic  limit  of  materials  was  but 
vaguely  recognized,  and  HOOKE'S  law  of  proportionality  of 
stress  to  elongation  was  often  applied  to  all  the  phenomena 
preceding  rupture.  YOUNG'S  statements  are  valid  in  a  general 
way,  but  we  now  know  that  it  is  necessary  to  distinguish  be- 
tween the  two  cases  of  elastic  resistance  and  non-elastic  resist- 
ance. Hence  there  are  two  divisions  of  the  subject  of  impact : 
first,  that  of  elastic  resilience,  where  the  molecular  forces^  do 
not  surpass  the  elastic  limit  of  the  material ;  and  second,  that 
of  ultimate  resilience,  where  the  elastic  limit  is  exceeded  and 
rupture  finally  occurs. 

The  problems  of  elastic  resilience  are  largely  theoretical  and 
mathematical.  Their  discussion  begins  with  YOUNG,  and  has 
been  continued  by  a  long  line  of  investigators  to  the  present 
time  ;  it  forms,  indeed,  the  most  prominent  part  of  the  theory 
of  elasticity,  which  has  been  so  thoroughly  set  forth  in  the 
history  of  TODHUNTER  and  PlERSON.  Starting  with  HOOKE'S 
law  of  proportionality  of  stress  to  deformation,  and  applying 
to  this  the  mechanical  laws  of  force,  velocity,  and  work,  the 
stresses,  displacements,  and  resilience  of  elastic  bodies  subject 
to  impact  have  been  deduced.  The  results  thus  theoretically 
found  have  been  confirmed  by  many  experiments,  as  must 
necessarily  be  the  case,  since  all  the  laws  of  the  discussion  are 
those  of  experiment  and  experience. 


210  THE   RESILIENCE   OF  MATERIALS.  CH.  IX. 

The  first  problems  of  elastic  resilience  were  those  of  bars  or 
rods  subject  to  the  longitudinal  impact  of  a  moving  weight. 
The  vibrations  of  stretched  wires  had  been  discussed  by 
EULER,  LAGRANGE,  and  others ;  but  NAVIER,  in  1823,  appears 
to  have  been  the  first  to  investigate  the  oscillations  and  maxi- 
mum stresses  in  a  horizontal  bar  due  to  a  weight  impinging  on 
its  end.  PONCELET,  in  1829,  treated  the  same  problem  for  a 
vertical  bar  subject  to  longitudinal  impact  by  a  falling  weight. 
These  discussions  showed  that  for  the  case  of  horizontal  im- 
pact the  maximum  elongation  is  a  mean  proportional  between 
twice  the  height  of  fall  and  the  extension  due  to  the  same 
weight  when  applied  gradually,  while  for  vertical  impact  it  is 
the  sum  of  the  static  extension  and  the  hypothenuse  of  a 
right-angled  triangle,  whose  sides  are  this  static  extension  and 
the  dynamic  extension  stated  for  the  horizontal  bar.  It  is  a 
corollary  from  this  that  under  the  sudden  application  of  a  load, 
which  is  the  particular  case  of  impact  when  the  height  of  fall 
is  zero,  the  maximum  elongation,  and  hence  the  maximum  in- 
ternal stress,  is  twice  as  great  as  that  caused  by  the  same  load 
when  applied  gradually.  (Art.  93.) 

While  speaking  of  PONCELET,  it  is  well  to  pause  to  remark, 
that  his  work  on  industrial  mechanics  which  contains  these  in- 
vestigations deserves  especial  mention  as  the  book  that  marks 
the  beginning  of  technical  education  ;  his  conceptions  and  ex- 
planations of  these  problems  of  external  work  and  internal 
resilience  may  indeed  still  be  regarded  as  models  of  clear  and 
accurate  reasoning. 

The  solutions  of  NAVIER  and  PONCELET  for  longitudinal 
impact  on  a  bar  were  approximate  in  the  sense  that  its  weight 
was  regarded  as  small  compared  to  that  of  the  striking, body. 
Later  investigations  by  SAINT  VENANT,  BOUISSINESQ,  and 
others,  have  completely  resolved  the  problem  when  the  rela- 
tive masses  are  taken  into  account,  and  have  also  disclosed  all 


ART.  98.  EARLY  HISTORY  OF  RESILIENCE.  211 

the  attendant  circumstances  of  duration  and  intensity  of  the 
forces  at  the  surface  of  impact. 

Another  important  problem  is  that  of  transverse  impact  on 
a  beam  giving  rise  to  flexural  resilience.  The  first  investiga- 
tion of  this  was  by  HODGKINSON,  1833,  in  connection  with  the 
discussion  of  experiments,  while  the  full  elastic  theory  for  dif- 
ferent ratios  of  weights  of  the  beam  and  falling  body  was  de- 
veloped by  Cox  in  1848,  and  later  most  fully  extended  by  the 
French  elasticians  in  a  similar  manner  to  that  of  longitudinal 
impact.  The  law  of  proportionality  of  resilience  to  volume 
was  shown  to  be  true  only  when  the  latter  is  increased  an 
amount  proportional  to  one-half  the  ratio  of  the  weight  of 
the  beam  to  that  of  the  striking  body,  and  similar  modifications 
are  necessary  regarding  deflections  and  internal  stresses.  In 
many  problems  the  velocity  of  transmission  of  stress,  which 
is  the  same  as  the  velocity  of  sound  in  the  given  material, 
forms  an  important  element,  as  indeed  YOUNG  had  suggested 
in  1807,  at  the  very  beginning  of  the  consideration  of  impact 
resistances. 

While  the  conclusions  of  all  these  investigations  of  elastic 
resilience  are  true  and  valuable,  it  should  never  be  forgotten 
that  they  are  only  true  when  the  conditions  are  observed  under 
which  they  are  deduced,  namely,  that  the  elastic  limit  is  not 
exceeded  by  the  maximum  internal  stress.  Now,  while  the 
elastic  limit  for  some  materials  is  as  high  as  one-half  the  ulti- 
mate strength,  the  elongation  up  to  the  elastic  limit  is  small 
compared  with  the  ultimate  elongation ;  and  hence  the  elastic 
resilience  may  be  very  small — less  perhaps  than  one  part  in  a 
thousand,  compared  with  the  total  ultimate  resilience.  The 
phenomena  of  elastic  resilience  form,  indeed,  such  a  small  por- 
tion of  those  occurring  in  practice,  that  it  is  difficult  to  observe 
them  with  precision,  while  those  of  non-elastic  resilience  are 
ever  present.  Many  a  misconception  has  arisen  and  many  a 


212  THE  RESILIENCE   OF   MATERIALS.  CH.  IX 

paradox  has  been  founded  on  the  assumption  that  the  theory 
of  the  smaller  part  is  applicable  also  to  the  larger  part  of  the 
phenomena  (Art.  loo). 

The  investigation  of  ultimate  resilience  is  necessarily  experi- 
mental, since  beyond  the  elastic  limit  no  theoretical  relation 
between  stress  and  deformation  is  known.  In  1818  TREDGOLD 
made  experiments  on  wooden  beams  subject  to  the  impact  of 
a  falling  ball,  and  concluded  that  YOUNG'S  law  of  proportion, 
ality  of  resilience  to  volume  was  not  justified.  HODGKI:;SON, 
in  1835,  experimented  on  cast-iron  beams  under  lateral  impact 
from  a  pendulum,  and  found  that  the  deflections  were  propor- 
tional to  the  velocities  of  impact,  that  the  same  work  was 
required  to  break  the  beam  whether  struck  at  the  middle  or  at 
the  quarter  point,  and  that  the  weight  of  the  beam  increased 
its  ultimate  resilience — all  of  which  is  more  nearly  in  accord- 
ance with  the  elastic  theory  than  perhaps  might  be  expected. 
In  1849  was  published  the  •'  Report  of  the  Commissioners 
appointed  to  inquire  into  the  application  of  iron  to  Railway 
Structures,"  which  contains  some  of  the  most  comprehensive 
and  valuable  experiments  ever  undertaken.  In  view  of  later 
practice  it  was  perhaps  unfortunate  that  these  were  mostly 
made  upon  cast  iron,  but  the  methods  of  investigation  and  the 
conclusions  deduced  render  this  report  an  epoch-making  one 
in  the  history  of  the  resistance  of  materials.  Here  can  only 
be  noted  those  points  relating  to  impact  on  which  experiments 
were  made  by  HODGKINSON,  WILLIS,  and  GALTON.  Trans- 
verse impact  on  beams  by  oft-repeated  blows  of  a  ball  swung 
as  a  pendulum,  and  also  by  pressure  applied  by  a  revolving 
cam,  indicated  for  the  first  time  the*  laws  of  fatigue  under  re- 
peated stresses ;  transverse  impact  with  a  single  blow  confirmed 
YOUNG'S  theorem  of  resilience  and  volume,  as  the  same  amount 
of  work  was  required  to  rupture  a  rectangular  beam  whether 
struck  on  its  narrow  or  broad  side ;  and  the  inertia  of  the  beajn 
was  shown  to  be  an  important  factor  in  increasing  its  resilience. 


ART.  99.  MODERN   EXPERIMENTS.  213 

Weights  suddenly  applied  without  impact  gave  deflections 
nearly  double  the  static  deflections,  and  the  same  ratio  was 
observed  in  the  breaking  loads.  Experiments  on  the  effects 
of  a  load  moving  with  different  velocities  over  a  beam  showed 
that  the  deflection  could  amount  to  more  than  double  the 
static  deflection,  and  the  elaborate  theoretical  discussions  of 
this  case  by  STOKES,  together  with  the  evidence  given  by  ex- 
perienced engineers,  form  the  basis  of  many  practical  rules 
since  used  in  bridge  design.  The  immediate  result  indeed  was 
the  revision  of  the  factors  of  safety  by  the  British  Board  of 
Trade  and  the  establishment  of  a  rule  that  for  cast  iron  the 
factor  for  live  load  should  be  double  that  for  dead  load. 

ART.  99.    MODERN  EXPERIMENTS. 

The  thirty  years  following  the  middle  of  the  century  may  be 
designated  as  the  period  of  development  of  the  modern  methods 
of  static  tests  upon  tensile  specimens.  In  the  United  States 
this  development,  begun  by  WADE,  RODMAN,  and  PLYMPTON, 
later  produced  the  numerous  testing  machines  of  FAIRBANKS, 
OLSEN,  and  RIEHLE,  and  at  last  has  culminated  in  the  precise 
apparatus  of  EMERY  and  in  the  powerful  machines  at  Athens 
and  Phcenixville.  These  static  tensile  tests  seem  to  have 
diverted  attention  from  the  question  of  impact  and  to  have 
caused  the  neglect  of  the  theory  of  resilience,  for  the  objects 
of  the  method  have  been  mainly  to  determine  the  elastic  limit, 
maximum  strength,  and  ultimate  elongation,  considerations  of 
work  having  been  largely  overlooked.  Yet  the  diagram  formed 
by  laying  off  the  unit-stresses  as  ordinates  and  the  correspond- 
ing elongation  as  abscissas  shows  by  its  area  the  internal  work 
which  resists  rupture,  and  thus  is  approximately  a  measure  of 
the  ultimate  resilience  per  cubic  unit  of  the  material.  The 
phenomena  in  a  slow  test,  however,  are  not  in  all  respects  the 
same  as  in  a  rapid  one,  the  elastic  limit  and  maximum  strength 
being  generally  greater  in  the  latter,  while  the  elongation  is 


214  THE   RESILIENCE  OF  MATERIALS.  CH.  IX. 

less.  How  the  diagram  would  look  under  the  very  rapid  de- 
velopment of  stress  is  yet  to  be  determined,  and  not  until  this 
is  done  by  an  autographic  device  can  a  full  knowledge  of  the 
total  field  of  resilience  be  obtained.  High  values  of  the  ulti- 
mate elongation  and  contraction  of  area  have  been  regarded  as 
of  great  importance  in  judging  of  the  quality  of  iron  and  steel, 
and  this  has  given  rise  to  what  may  be  called  an  unconscious 
manipulation  of  the  machine,  the  power  being  applied  at  such 
a  rate  as  to  secure  these  results  consistent  with  other  condi- 
tions. It  is  safe  to  say  that  stresses  are  not  applied  and  that 
rupture  does  not  occur  in  this  manner  in  practice  ;  and  here  it 
is  that  static  tests  fail  to  give  full  satisfaction. 

It  would  be  easy  to  take  a  great  deal  of  space  in  criticism  of 
the  imperfections  of  the  static  method  of  testing,  to  show  the 
uncertainties  of  percentages  of  elongation,  to  dwell  on  the  dis- 
crepant results  found  from  specimens  of  different  shape,  and 
to  state  the  various  views  regarding  contraction  of  area  and 
the  appearance  of  the  fracture.  But  this  is  not  the  place.  It 
may  be  said  again,  however,  that  the  work  per  cubic  unit  de- 
veloped before  rupture  is  an  important  element,  by  the  study 
of  which  much  can  be  learned ;  for  the  total  resilience  depends 
not  merely  on  maximum  strength  and  ultimate  elongation,  but 
upon  the  rate  of  variation  of  these  throughout  the  test.  Each 
material,  or  specimen,  by  virtue  of  its  physical  and  chemical 
properties  and  its  method  of  manufacture,  has  the  capacity  for 
resisting  a  certain  amount  of  applied  work,  of  which  the  quanti- 
ties measured  in  the  static  test  are  not  complete  factors.  The 
full  resilience  can  indeed  only  be  found  from  an  autographic 
diagram  like  that  used  in  THURSTON's  torsion  machine,  which 
has  done  good  service  in  this  direction. 

During  all  this  development  of  static  testing  one  impact  test 
has  survived  and  everywhere  held  its  own.  This  is  the  cold- 
bend  test  for  wrought  iron  and  steel.  In  the  rolling-mill  it  is 


ART.  99.  MODERN   EXPERIMENTS.  21$ 

used  to  judge  of  the  purity  and  quality  of  the  muck  bar ;  in 
the  steel  mill  it  serves  to  classify  and  grade  the  material  almost 
as  well  as  chemical  analysis  can  do,  and  in  the  purchase  of 
shape  iron  it  affords  a  quick  and  reliable  method  of  estimating 
toughness,  ductility,  strength,  and  resilience.  It  is  true  that 
numerical  values  of  these  qualities  are  not  obtained,  but  the 
indications  are  so  valuable,  that  if  all  tests  except  one  were  to 
be  abandoned,  the  simple  cold-bend  test  would  probably  be 
the  one  which  the  majority  of  engineers  would  desire  to  retain. 

Returning  now  to  impact  experiments  proper,  the  experi- 
ments of  KlRKALDY,  in  1862,  deserve  notice.  He  subjected 
wrought-iron  specimens  to  sudden  tensile  stress  by  loads  which 
were  applied  without  velocity,  and  found  that  they  broke  with 
82  per  cent  less  load  than  under  static  conditions.  This  has 
since  been  quoted  as  a  contradiction  of  the  law  that  the  stress 
produced  by  a  gradual  load  is  one-half  that  caused  by  the  same 
load  when  suddenly  applied  ;  but  really  no  contradiction  exists, 
since  the  law  is  only  true  when  the  elastic  limit  is  not  exceeded. 
For  ultimate  resilience  the  ratio  of  gradual  to  sudden  load  pro- 
ducing the  same  stress  is  always  greater  than  one-half,  and 
approaches  nearer  to  unity  the  more  ductile  the  material. 
KlRKALDY  also  introduced  the  ultimate  contraction  of  area  as 
an  element  of  equal  importance  with  the  ultimate  elongation  ; 
and  undoubtedly  this  is  the  case,  since  it  is  subject  to  less 
variation  with  the  length  of  the  specimen.  However,  little 
satisfactory  evidence  exists  that  contraction  of  area  is  a  meas- 
ure of  resistance  to  impact,  as  some  have  supposed. 

The  impact  hammer  or  ram  delivering  repeated  blows,  as 
used  by  SANDBERG  and  STYFFE  in  1868  on  railroad  rails  and 
by  the  United  States  Board  for  Testing  Metal  in  1878  in  ex- 
periments on  chain  iron,  is  perhaps  the  most  common  method 
of  conducting  tests  for  resilience.  The  weight  of  the  ram 
multiplied  by  the  height  of  fall  measures  the  work  done  in  one 


2l6  THE   RESILIENCE   OF   MATERIALS.  CH.  IX. 

blow,  and  the  number  of  foot-pounds  required  to  produce  rup- 
ture furnishes  an  index  of  the  ultimate  resilience  of  the  bar  or 
specimen.  Although  valuable  for  comparative  purposes,  like 
the  cold-bend  test,  this  method  fails  to  give  proper  numerical 
results,  since  the  effect  of  a  second  blow  is  not  really  added  to 
that  of  the  first  on  account  of  the  dissipation  of  internal  work 
into  heat  which  follows  the  release  of  the  stress  when  this  is 
greater  than  the  elastic  limit.  That  the  work  of  several  blows 
affords  little  indication  of  ultimate  resilience  was  well  shown 
by  UCHATIUS,  in  1874,  who  experimented  on  rods  of  small 
size  under  the  impact  of  a  small  ram,  and  found  that  rupture 
could  be  caused  by  one  blow  of  28  inches  fall  giving  6  foot- 
pounds, by  four  blows  of  21  inches  fall  giving  18  foot-pounds,  by 
37  blows  of  loj  inches  fall  giving  83  foot-pounds,  or  finally  by 
2  050  blows  of  3J  inches  fall  giving  I  536  foot-pounds  of  work. 
Indeed,  the  fact  that  one  strong  blow  is  more  efficient  than 
many  weak  ones  is  well  known  to  every  artisan.  The  compar- 
ative weights  of  the  ram  and  bar  are  also  of  importance  in 
such  tests,  as  will  be  later  noticed. 

The  rise  of  the  elastic  limit  after  the  release  of  a  stress 
which  surpasses  it,  first  shown  by  THURSTON  in  1873  >an^  the 
consequent  stiffening  of  the  material  under  several  such  appli- 
cations of  stress,  is  also  observed  in  cases  of  repeated  impact. 
Thus  MARTENS,  in  1879,  tested  a  railroad  rail  by  a  ram  falling 
alternately  on  the  head  and  base ;  the  second  blow  produced 
only  95  per  cent  of  the  deflection  caused  by  the  first,  and  after 
four  or  five  blows  only  90  per  cent  was  obtained.  Here,  of 
course,  a  large  part  of  the  work  of  each  blow  was  dissipated  in 
heat  due  to  the  change  of  molecular  structure,  so  that  the  total 
work  expended  gives  no  valid  numerical  measure  of  the  true 
ultimate  resilience  of  the  material.  Our  knowledge  as  to  what 
occurs  under  such  repeated  impacts  is  of  the  most  uncertain 
kind.  Most  of  the  stresses  are  no  doubt  injurious,  as  the  final 
result  proves,  but  a  few  may  perhaps  be  beneficial  in  improv- 


ART.  99.  MODERN   EXPERIMENTS.  2I/ 

ing  the  quality  of  the  material  like  the  process  of  cold-hammer- 
ing and  cold-rolling  in  manufacture.  BAUSCHINGER'S  discovery 
in  1885,  that  the  raising  of  the  tensile  elastic  limit  is  accom- 
panied by  a  lowering  of  the  compressive  elastic  limit,  is  a  most 
important  one,  which  is  likely  in  connection  with  future  inves- 
tigations to  lead  to  a  clearer  knowledge  of  the  laws  of  stress  in 
materials  under  repeated  impact. 

The  investigations  of  WoHLER  between  1860  and  1870,  and 
the  later  ones  of  SPANGENBERG,  on  the  resistance  of  materials 
under  repeated  stresses  applied  with  little  impact,  have  been 
of  the  greatest  value  in  influencing  the  rational  design  of 
bridges  and  other  structures  subject  to  variations  of  load. 
They  establish  the  facts  that  repeated  stresses  below  the  elas- 
tic limit  do  not  injure  the  material,  that  repeated  stresses  be- 
yond  this  limit  cause  injury  in  proportion  to  the  range  between 
the  maximum  and  minimum  limits,  and  that  the  greater  the 
range  the  less  the  number  of  repetitions  required  to  produce 
rupture.  (Art.  91.)  The  prompt  adoption  of  rules  for  design- 
ing based  on  these  conclusions,  in  place  of  the  arbitrary  methods 
of  adding  20  or  30  per  cent  to  the  static  stresses,  indicated  that 
the  engineering  profession  appreciated  the  importance  of  pro- 
viding for  the  resistance  to  impact  in  a  rational  way.  Yet  prob- 
ably greater  impacts  occur  on  bridges  than  these  repeated 
stresses,  for  failures  are  not  infrequent,  and  the  life  of  a  bridge 
under  heavy  traffic  scarcely  exceeds  a  dozen  years.  The  fur- 
ther study  of  the  effect  of  repeated  impact  is  thus  imperative, 
for  what  we  now  know  is  but  little  compared  to  what  is  to  be 
learned. 

MAITLAND,  in  1887,  showed  by  many  experiments  on  tensile 
specimens  subject  to  many  blows  of  a  falling  ram  that  the  ulti- 
mate elongation  was  greater  than  in  static  tests,  it  being  in 
some  cases  nearly  doubled.  This  might,  perhaps,  be  expected 
from  what  has  been  said  regarding  the  beneficial  influence  of 


2l8  THE   RESILIENCE   OF   MATERIALS.  CH.  IX. 

some  of  the  stresses.  He  also  exploded  powder  and  gun-cot- 
ton between  two  cylinder  heads  joined  together  by  rods,  and 
found  their  ultimate  elongation  to  be  more  than  double  the 
corresponding  ones  under  static  stress.  This,  on  the  other 
hand,  might  not  have  been  expected  from  the  general  conclu- 
sions regarding  the  effect  of  time  on  stress  and  elongation,  and 
it  seems  not  to  be  the  case  for  the  impact  of  a  single  blow  in 
later  experiments.  Thus  it  appears  in  problems  of  ultimate 
resilience,  where  no  theory  has  yet  been  developed,  that  gen- 
eral principles  derived  from  static  tests  should  be  applied  with 
caution. 

The  tests  made  by  E.  D.  ESTRADA,  in  1893,  have  recently 
led  to  much  interesting  discussion  on  the  subject  of  impact. 
Specimens  of  wrought  iron  and  steel  were  tested  by  the  usual 
tension  machine,  and  also  under  the  blow  of  a  ram  weighing 
100  pounds  falling  through  vertical  heights  ranging  from  5  to 
25  feet.  Over  40  specimens  were  broken  under  repeated 
blows,  they  being  so  arranged  that  they  were  brought  into  ten- 
sion by  the  pressure  caused  by  the  impact.  The  number  of 
blows  ranged  from  2  to  14,  the  height  of  fall  varying  in  differ- 
ent cases.  These  valuable  experiments  show  clearly  that  the 
ultimate  elongation  under  repeated  impact  is  greater  than  for 
static  stresses,  the  average  of  all  being  about  one-third  greater, 
while  the  contraction  of  area  shows  only  a  small  increase  ;  but 
no  conclusions  regarding  the  elastic  limit  or  the  maximum 
strength  under  impact  can  be  derived  from  them.  The  impact 
was  not  directly  applied  to  the  specimen,  but  through  a  num- 
ber of  plates  and  bolts  through  which  it  was  transmitted,  and 
thus  much  of  the  work  was  spent  in  acting  against  their  inertia 
and  resilience.  This,  however,  in  no  way  invalidates  the  im- 
portant deductions  regarding  the  influence  of  repeated  impact 
on  ultimate  elongation  and  contraction  of  area,  and  the  con- 
clusion of  KlRKALDY  regarding  the  value  of  the  latter  as  an 
index  of  toughness  and  ductility  is  thoroughly  confirmed. 


ART.  99.  MODERN   EXPERIMENTS.  2 19 

The  elongation  before  rupture  appears  to  vary  approximately 
as  the  amount  of  work  expended  by  the  ram,  but  the  work  re- 
quired to  produce  rupture  does  not  give  a  satisfactory  com- 
parison of  the  ultimate  resilience  of  different  specimens,  except 
in  those  cases  where  the  height  of  fall  was  the  same. 

The  experimental  work  thus  briefly  reviewed  has  been  done 
by  special  improvised  apparatus,  and  with  little  or  no  uniform- 
ity of  method  ;  but  the  time  may  come  when  machines  for 
impact  tests  will  be  put  on  the  market  and  be  in  common  use. 
At  present  almost  the  only  one  that  can  be  mentioned  is  that 
devised  by  W.  J.  KEEP,  about  1889,  for  testing  jthe  resilience 
of  cast  iron.  The  blows  are  delivered  by  a  hammer  weighing 
25  pounds,  falling  like  a  pendulum  through  heights  less  than  6 
inches,  and  produce  horizontal  flexure  at  the  middle  of  a  small 
bar  i  foot  in  length.  Beginning  with  a  fall  of  £  inch,  successive 
blows  are  applied,  each  with  a  fall  -J  inch  greater  than  the  pre- 
ceding, until  rupture  occurs.  The  deflections  and  sets  of  the 
bars  are  graphically  recorded  by  the  machine  itself,  and  thus 
excellent  comparisons  of  the  ultimate  resilience  of  different 
grades  of  metal  may  be  obtained. 

Impact  tests  are  most  important  in  the  case  of  railroad  rails, 
car  wheels,  tires  and  axles,  and  other  forms  liable  to  shock. 
Such  rail  tests  have  been  carried  on  for  many  years  in  Europe, 
and  since  1890  in  the  United  States  by  P.  H.  DUDLEY. 
Already  at  least  three  prominent  railroads  require  drop  tests  of 
steel  rails  to  be  made  at  the  mill.  One  rail  butt  4^  feet  long  is 
to  be  taken  from  each  heat,  placed  on  solid  supports  with  either 
head  or  base  upwards,  and  a  weight  of  2000  pounds  is  dropped 
upon  it,  the  height  of  fall  being  16  feet  for  rails  lighter  than 
70  pounds  per  yard  and  20  feet  for  heavier  ones,  while  the  dis- 
tance between  the  supports  is  3  feet  for  the  former  and  4  feet 
for  the  latter.  Under  this  test  90  per  cent  of  the  rails  must 
not  break,  and  the  elongation  of  the  base  or  head  under  the 


220  THE  RESILIENCE  OF   MATERIALS.  CH.  IX. 

greatest  tension  must  be  more  than  5  per  cent.  This  require- 
ment of  testing  by  a  single  blow  is  in  every  respect  more  satis- 
factory than  by  several  repeated  ones,  as  the  deflection  and  de- 
formation are  produced  by  a  known  quantity  of  work  and  the 
complex  phenomena  of  stiffening  and  the  loss  of  work  in  heat 
are  largely  avoided. 

Aside  from  the  physical  qualities  of  the  metal,  impact  tests 
on  rails  and  wheels  promise  to  give  important  conclusions  re- 
garding the  influence  of  temperature,  of  chemical  composition, 
and  of  methods  of  manufacture,  and  thus  to  lead  to  a  better, 
more  uniform,  and  cheaper  product.  The  discovery  by  GOSS, 
in  1 892,  that  the  driving  wheels  of  a  locomotive  lift  up  from  the 
rail  during  a  part  of  each  revolution  when  moving  at  high 
speed  shows  that  impacts  are  of  more  common  occurrence  than 
generally  supposed,  and  to  satisfactorily  resist  these  an  in- 
creased resilience  in  wheels,  rails,  and  bridges  is  required. 

At  the  closing  session  of  the  Engineering  Congress  held  in 
1893  in  Chicago,  a  resolution  was  offered  by  DEBRAY  that  uni- 
form methods  of  testing  are  desirable  for  purposes  of  compari- 
son, and  it  was  adopted  unanimously.  For  common  static 
tests  of  tension  and  compression  the  time  has  certainly  arrived 
when  rules  to  secure  uniformity  should  be  framed  and  followed. 
The  conferences  held  at  Munich  in  1884  and  at  Dresden  in 
1886  made  a  good  beginning,  and  the  later  work  done  in  1891 
by  the  Committee  of  the  American  Society  of  Mechanical 
Engineers  will  undoubtedly  bear  good  fruit.  With  respect  to 
impact,  however,  our  knowledge  is  not  yet  sufficient  to  frame 
uniform  rules.  The  recommendation  that  the  weight  of  the 
anvil,  or  supporting  blocks  which  hold  the  specimen,  should  be 
at  least  ten  times  that  of  the  ram  is  an  excellent  one,  but 
others  fully  as  important  will  doubtless  be  developed  by  fur- 
ther rational  investigation  and  experiment. 

The  influence  of  the  inertia  of  the  resisting  body  in  modify- 


ART.  99.  MODERN   EXPERIMENTS.  221 

ing  the  effect  of  the  impact  should  not  be  forgotten.  If  the 
ram  be  light,  local  damage  in  the  body  struck  will  be  the  re- 
sult rather  than  the  development  of  its  resilience.  COTTERILL 
has  suggested  for  the  case  of  transverse  impact  that  if  the  total 
work  of  the  ram  be  taken  as  proportional  to  its  weight  plus 
one-half  the  weight  of  the  beam,  then  the  work  spent  in  de- 
veloping resilience  may  be  taken  as  proportional  to  the  weight 
of  the  ram  and  that  spent  in  local  damage  to  one-half  the 
weight  of  the  beam.  This  is  a  good  rule  to  keep  in  mind,  for 
longitudinal  impact  producing  pure  tension,  one-third  being 
the  fraction  to  be  used  instead  of  one-half.  The  fall  of  the 
ram  should  not  be  too  great,  for  a  high  velocity  of  impact  is 
also  apt  to  cause  the  work  to  be  spent  locally,  since  time  is  re- 
quired for  the  transmission  of  internal  stress.  The  cutting  of 
hard  steel  plates  by  the  impact  of  particles  of  sand  at  high 
velocities  is  an  example  familiar  to  engineers,  and  that  drops 
of  water  will  wear  away  stone  is  known  to  all ;  hence  a  low 
fall  combined  with  a  heavy  ram  seems  best  adapted  for  con- 
ducting impact  tests  where  the  most  satisfactory  numerical 
results  are  desired. 

It  may  further  be  suggested  that  the  blow  should  be  de- 
livered directly  to  the  beam  or  specimen  without  the  interven- 
tion of  intermediate  blocks  or  parts  which  may  absorb  work. 
At  the  same  time  the  surface  of  contact  should  be  sufficiently 
large  so  that  the  compression  of  the  ram  may  not,  if  possible, 
exceed  the  elastic  limit,  and  thus  loss  of  work  in  heat  be 
avoided.  In  elastic  resilience  the  applied  work  is  not  trans- 
formed into  heat,  in  resilience  accompanied  by  permanent  de- 
formation it  is ;  and  hence  the  tests  should  be  so  conducted 
that  the  specimen  and  not  the  ram  may  be  heated.  Lastly,  it 
may  be  noted  that  the  rebound  of  the  ram  should  be  subtracted 
from  the  total  fall  to  give  an  exact  measure  of  the  work 
actually  performed  by  it  upon  the  body  which  is  tested. 


222  TENSION  AND   COMPRESSION.  CH.  X. 


CHAPTER    X. 
TENSION   AND   COMPRESSION. 

ART.   100.     ELONGATION  UNDER  OWN  WEIGHT. 

When  an  unloaded  bar  is  hung  vertically  by  one  end  there 
is  no  stress  on  the  lower  end,  while  at  the  upper  end  there  is  a 
stress  equal  to  the  weight  of  the  bar.  In  many  practical  cases 
this  stress  is  so  small  that  it  can  be  neglected  in  comparison 
with  that  caused  by  the  applied  tension. 

The  elongation  of  a  vertical  bar  under  its  own  weight  is  one- 
half  that  caused  by  the  same  load  applied  at  the  end.  To 
show  this,  let  /  be  the  length  of  the  bar,  A  the  area  of  the 
cross-section,  and  W  the  weight  ;  let  x  be  any  distance  from 

x 
the  lower  end,  then  Wj  is  the  weight  of  this  portion.     The 

elementary  elongation  caused  by  it  on  the  length  dx  is,  from 
Art.  5, 

Wxdx 


and  the  integral  of  this  between  the  limits  o  and  /  gives 

Wl 

A'  ==  ~2~AE> 
which  is  one-half  of  that  caused  by  a  load  W  at  the  end. 

The  work  performed  by  gravity  in  elongating  the  bar  is  one- 
third  of  that  done  by  the  same  load  applied  at  the  end.  For, 
the  elementary  work  done  upon  the  element  dx  by  the  lower 
portion  of  the  bar  is 

\Wx  Wx^dx 

~2~Td^  -    2AEI*  ' 


ART.   101.  BAR   OF   UNIFORM   STRENGTH.  223 

the  integral  of  which,  between  the  limits  o  and  /,  gives 


while  that  done  by  Wat  the  end  of  the  bar  is  %W\  or 

The  total  elongation  caused  by  the  weight  W  and  a  load  P 
at  the  end  hence  is 


and  the  total  work  of  elongation  is 


These  formulas,  let  it  be  remembered,  are  only  valid  when  the 
unit-stress  (P  +  W)/A  is  less  than  the  elastic  limit  of  the 
material.  They  apply  also  to  the  case  of  compression  within 
the  elastic  limit,  A.  being  the  amount  of  shortening. 

Prob.  144.  Find  the  length  of  a  wrought-iron  bar,  and  its 
elongation,  when  suspended  at  the  upper  end  so  that  the  unit- 
stress  at  that  end  may  be  equal  to  the  elastic  limit. 

ART.  10  1.     BAR  OF  UNIFORM  STRENGTH. 

A  suspension  bar  of  constant  cross-section  is  stressed  at  the 
lower  end  by  the  load  P,  and  at  the  upper  end  by  P  plus  the 
weight  of  the  rod.  When  the  bar  is  very  long  and  heavy  its 
section  should  be  less  at  the  lower  than  at  the  upper  end  in 
order  to  economize  material.  The  bar  in  such  cases  is  some- 
times made  in  parts,  the  cross-section  of  any  part  being  less 
than  that  of  the  one  above  it. 

The  theoretic  form  for  a  bar  of  uniform  strength  may  be 
determined  as  follows  :  Let  P  be  the  load  applied  to  the 
lower  end,  and  let  5  be  the  allowable  working  unit-stress. 
Then  the  area  of  the  lower  end  is 


224  TENSION  AND   COMPRESSION.  CH.  X. 

Let  A  be  the  area  of  any  section  at  a  distance  y  from  the 
lower  end,  then  A  +  dA  will  be  the  area  at  the  distance 
y  +  dy,  and  the  area  dA  must  provide  for  the 

weight  in  the  distance  dy.     Let  w  be  the  weight 

per  cubic  unit  of  the  bar,  then 

dA  =  ^, 

from  which  by  integration,  and  observing  that 
A  =  A,  when  y  =  o,  there  is  found 

S* 

iv 

Replacing  for  S  its  value  P/A9,  this  may  be 


FIG.  57. 

written 


wA 
=  log, -4.  +  --y, 


where  the  logarithms  are  in  the  Naperian  system.     Passing  to 
common  logarithms,  it  becomes 


log  A  =  log  A.  +  0.43429  -ply, 
which  is  the  form  for  practical  computation. 

For  example,  let  P  =  10000  pounds,  S  =  $  ooo  pounds  per 
square  inch,  and  w  =  0.25  pounds  per  cubic  inch.  Then  A9 
=  2  square  inches,  and  the  formula  becomes 

log  A  =  0.30103  +  O.OOOO2I7/. 

Now  for  y  =  100  inches,  log  A  =  0.30320,  and  A  =  2.01  square 
inches;  for  y  =  1000  inches,  log  A  =  0.32274  and  A  =2.10 
square  inches  ;  while  when/  =  lOOOO  inches,  A  —  3.30  square 
inches.  Thus  it  is  only  in  the  case  of  very  long  bars  that  an 
appreciable  increase  in  cross-section  is  found. 

Prob.  145.  Let  a  pier  whose  top  section  is  a  rectangle  of 
length  /  and  breadth  b  support  the  load  P,  as  in  Prob.  26. 
Deduce  the  value  of  log  x  in  terms  of  b,  /,  and  P. 


ART.  102.  LONGITUDINAL  IMPACT.'  22$ 

ART.  102.     LONGITUDINAL  IMPACT. 

By  longitudinal  impact  is  understood  the  impinging  of  a 
moving  body  upon  the  end  of  a  bar  so  as  to  produce  in  it 
either  tensile  or  compressive  stress.  In  Art.  93  this  subject 
was  discussed  and  formulas  were  deduced  for  the  elongation 
and  stress  under  impact.  These  formulas,  however,  take  no 
account  of  the  resisting  influence  of  the  inertia  of  the  bar, 
which  may  modify  the  results  when  the  weight  of  the  bar  is 
large  compared  to  that  of  the  moving  body. 

Let  a  body  of  weight  P  be  free  to  move  along  a  horizontal 
bar,  and  let  v  be  its  velocity  when  it  reaches  the  end  of  the 


FIG.  58. 

tf 
bar.    The  work  then  stored  in  it  is  P— }  or  Ph,  if  h  be  the 

height  due  to  the  velocity  v.  This  work  is  expended  in  over- 
coming the  inertia  of  the  particles  of  the  bar  and  in  elongating 
it  through  a  distance  y.  A  series  of  oscillations  then  results 
until  finally  the  end  of  the  bar  comes  to  rest  at  its  original 
position,  provided  that  the  elastic  limit  be  not  exceeded  by  the 
maximum  stress  produced. 

The  load  P  has  the  velocity  v  before  it  strikes  the  end  of 
the  bar.  When  in  complete  contact  both  P  and  the  end  of 
the  bar  are  moving  with  the  velocity  V  which  is  less  than  v. 
At  this  instant  any  element  of  the  bar  dW\s  moving  with  a 
velocity  u.  Thus  the  work  stored  up  by  P  and  the  bar  at  this 
instant  of  complete  contact  is 


Fa         Cl         u9 
K  =  P—  +    I   dW.—. 

**          Jo  2S 


226  TENSION  AND  COMPRESSION.  CH.  X. 

Now  u  —  o  for  the  fixed  end,  and  u  =  V  for  the  free  end  of 
the  bar,  and  in  general  u  is  proportional  to  the  distance  x  from 
the  fixed  end.  Thus 


and 
and  introducing  these  values,  the  integral  in  the  above  expres- 


sion  is  found  to  be  \W  —  ,  or  one  third  of  that  which  would 
obtain  if  the  entire  bar  were  in  motion.     Then 


is  the  work  stored  up  by  load  and  bar  when  the  end  of  the  bar 
and  the  load  are  moving  with  the  common  velocity  V. 

Now  at  this  instant  the  impulse  of  the  two  bodies  is  equal  to 
the  impulse  of  P  before  striking,  whence 

Pv 

or         F=/>  +  ^» 
and  hence  the  work  K  is 

P*         v*  Pv*  Ph 

~  P  +  ^W  2g  "  2g(i  +  P)  ==  i  +  \V 

in  which  k  denotes  the  ratio  of  W  to  P.  This  work  is  ex- 
pended in  elongating  the  bar  through  the  distance  y,  the  inter- 
nal  stress  increasing  from  o  up  to  Qy  so  that  the  entire  internal 
work  is  \Qy.  Hence 

ifi^rpp 

is  the  equation  between  external  and  internal  work. 

If  the  elastic  limit  be  not  exceeded  the  forces  Q  and  P  are 
proportional  to  the  elongations  they  can  produce.  Let  \  be 
the  static  elongation  due  to  P.  Then  Q/P  =  y/\  and  the 
equation  gives 

~2hK~  I       2h 

y  = 


ART.   102,  LONGITUDINAL    IMPACT. 

from   which   the   elongation  y  and   the   stress   Q  are  deter- 
mined. 

The  static  elongation  \  due  to  a  load  P  and  a  cross-section 
A  is,  from  Art.  5, 


and  this  may  be  inserted  in  the  above  formulas  either  in  literal 
or  numerical  form.  The  formulas  apply  equally  well  to  the 
case  of  compression  where  the  load  P  impinges  upon  the  end 
of  the  bar,  provided  that  /  be  not  so  long  that  lateral  flexure 
occurs. 

If  the  bar  be  placed  in  a  vertical  position  the  additional 
work  Py  is  performed  while  the  load  descends  through  the  dis- 
tance /,  and  the  equation  of  work  is 

Ph 


which  leads  to  the  formulas 


(20) 


and  these  are  the  same  as  those  of  Art.  93  if  the  ratio  k  be 
made  equal  to  o. 

The  influence  of  the  weight  of  the  bar  is  hence  to  diminish 
the  elongation  and  stress  under  impact.  For  instance,  in  the 
case  of  the  horizontal  bar  let  5"  be  the  unit-stress  produced  if 

£ 
the  weight  of  the  bar  be  very  small  ;  then  —  /  =  is  the  unit- 

VI  +  & 

stress  when  the  weight  of  the  bar  is  k  times  that  of  the  load. 
Thus  if  W  =  P  the  unit-stress  is  0.8/5,  and  if  W  =  gP  the 
unit-stress  is  0.56";  but  these  high  values  of  k  are  unusual. 


228  TENSION  AND   COMPRESSION.  CH.  X. 

As  a  numerical  example,  let  it  be  required  to  find  the  stress 
produced  in  a  vertical  wrought-iron  bar,  one  square  inch  in 
section  and  18  feet  long  by  the  impact  of  a  body  weighing  30 
pounds  falling  through  a  height  of  one  foot.  Here  W  =  60 
pounds  and  k  =  2.  The  static  elongation  is 

PI 

A  =  ~r-   =  o.ooo  2592  inches. 


Then,  as  h  =  12  inches,  formulas  (20)  give 

y  =  o.ooo  2592  X  236.7  =  0.0614  inches, 
Q  =  30  X  236.7  =  7100  pounds, 

which  shows  that  very  high  stresses  may  be  produced  by  the 
impact  of  light  bodies  falling  through  moderate  heights. 

Prob.  146.  A  weight  of  60  pounds  moving  horizontally  im- 
pinges upon  the  end  of  a  bar  of  wrought  iron  2  inches  in  diam- 
eter and  12  feet  long.  Find  the  velocity  v  which  will  stress 
the  bar  up  to  the  elastic  limit. 

ART.  103.     OSCILLATIONS  AFTER  IMPACT. 

Referring  to  the  case  of  longitudinal  impact  shown  in  Fig. 
58',  it  is  clear  that  the  velocity  of  the  weight  P  decreases  after 
it  strikes  the  end  of  the  bar,  and  that  the  Velocity  at  any 
instant  is  a  function  of  the  corresponding  elongation  x.  When 
x  equals  the  final  elongation  y,  the  velocity  is  zero  ;  the  end 
of  the  bar  now  springs  back  so  that  the  velocity  becomes 
negative,  and  this  increases  numerically  until  x  =  o,  and  then 
decreases  until  it  becomes  zero  at  x  =  —  y  ;  the  oscillation  is 
next  performed  in  the  opposite  direction.  These  oscillations 
would  continue  indefinitely  were  it  not  for  resistances  of  fric- 
tion, but  owing  to  these  they  become  less  and  less  in  ampli- 
tude until  finally  the  bar  comes  to  rest. 

Let  vx  be  the  velocity  of  the  end  of  the  bar  when  the  elon- 
gation x  is  attained,  and  let  Qx  be  the  corresponding  stress  in 
the  bar.  The  kinetic  energy.of  the  moving  bar  and  weight 


ART.  103.  OSCILLATIONS  AFTER  IMPACT.  22Q 

then  equals  the  internal  work  still  to  be  performed  in  increas- 
ing x  to  y,  or 


Replacing  Q  by  P.y/^  and  Qx  by  P.  x/\,  this  becomes 


which  gives  the  velocity  of  the  end  of  the  bar  for  any  value 
of  x.      When  x  =  -\-  y  or  x  =  —  y  this  velocity  is  zero. 

To  find  the  time  in  which  the  oscillation  from  -\-  y  to  —  y 
is  performed,  let  /  be  the  time  counted  from  the  instant  when 
x  =  o.  Putting  vx  =  dx/dt,  the  last  equation  becomes 


dt  = 


g      vy  -  **' 

the  integration  of  which  gives 


-V 


When  ^r  =  y  the  arc  is  \n,  and  when  x  =  —  y  the  arc  is  f  TT. 
Hence  the  number  of  seconds  in  one  oscillation  is 


=V- 


which  is  the  same  as  the  time  of  oscillation  of  a  pendulum 
whose  length  is  A(i  +  \K). 

As  a  numerical  example  let  the  data  of  the  last  problem  be 
considered.  Here  P  =  60  pounds,  W=  125.7  pounds,  k  = 
2.09,  ^  =  0.00011  inches,  g  =  32.2  X  12  inches  per  second 
per  second.  Then  the  time  of  one  oscillation  is  t  =  0.022 


230  TENSION    AND    COMPRESSION.  CH.  X. 

seconds,  and  about  460  oscillations  would  be  performed  in  one 
second  were  it  not  for  frictional  resistances. 

For  the  case  of  the  vertical  bar  the  first  equation  of  this 
article  will  be  modified  by  subtracting  P(y  —  x)  from  the 
second  member,  this  expressing  the  potential  energy  to  be 
expended  by  the  weight  in  falling  through  the  distance^  —  x. 
The  expression  for  the  time  is  found  to  be 


t  =  \  /  -  — '  versin 


g  y  -  A' 

which  shows  that  the  middle  of  the  oscillation  occurs  at  x  =  A 
and  that  the  bar  finally  comes  to  rest  with  the  elongation  A. 
The  time  of  one  oscillation  is  the  same  as  before 

If  a  weight  W^  be  fixed  at  the  end  of  the  bar  and  the  weight 
P  impinge  upon  it,  then  the  formulas  of  this  and  the  last 
article  apply  if  I  +  t^  De  replaced  by  I  +  k, .  +  \k,  in  which 
k,  is  the  ratio  WJP. 

ART.  104.    CENTRIFUGAL  STRESS. 

When  a  body  of  weight  P  revolves  around  an  axis  with  the 
uniform  velocity  vy  and  r  is  its  distance  from  the  axis,  a  cen- 
trifugal force  Q  is  generated  whose  value,  as  deduced  in  theo- 
retical mechanics,  is 


and  which  acts  as  a  stress  in  the  cord  or  bar  that  connects  the 
body  with  the  axis.     The  case  shown  in  Fig.  59  is  that  of  a 


FIG.  59. 
bar  of  uniform  cross-section  and  length  /  having  a  weight  P 


ART.  104.  CENTRIFUGAL  STRESS.  231 

attached  to  one  end  while  it  revolves  around  an  axis  A  at  the 
other  end.  It  is  required  to  find  the  centrifugal  stress  in  the 
bar  at  A  when  the  speed  of  n  revolutions  per  second  is  main- 
tained. 

Let  x  be  any  distance  from  the  axis  •  the  velocity  at  this 
distance  is  2nxn,  or,  if  GO  be  the  angular  velocity,  the  velocity  at 

the  distance  x  is  XGO,  and 

oo 
GO  =  27tn,         or         n  =  — . 

Now  let  Wbe  the  weight  of  the  bar,  and  dW 'an  element  at 
the  distance  x.  Then  the  centrifugal  stress  at  A  is 


Q  =  P—  oo9  +    \    dW 

&  I/O 


g* 

But  dW  =  Wdx/l\  inserting  this  and  integrating, 


o 

which  gives  the  centrifugal  stress  at  the  axis. 

As  an  example  let  a  bar  of  wrought  iron  2X2  inches  and  6 
feet  long  have  a  weight  of  400  pounds  attached  at  6J  feet  from 
the  axis  of  revolution.  It  is  required  to  find  the  number  of 
revolutions  per  second  in  order  to  produce  rupture.  Solving 
the  last  equation  for  a?,  there  results, 

.          Qz 
-  Pr  +  twr 

in  which  Q  =  2  X  2  X  55  ooo  =  220000,  P  =  400  pounds, 
W  =  80  pounds,  g  =  32.16  feet  per  second  per  second,  /  =  6 
feet,  r  =  6  J  feet;  then  GO  =  50.8,  and 

50.8 
n  =  --  =8.1  revolutions  per  second, 

27T 

which  is  the  speed  required. 

Another  case  is  that  of  a  thin  circular  rim  or  hoop  of  mean 
radius  r  and  thickness  /  which  revolves  uniformly  around  its 


232  TENSION  AND   COMPRESSION.  CH.  X. 

center   as  an   axis.     Let  W  be   its  weight,  which  is   closely 
equal  to  2nrtwy  if  w  be  the  weight  of  the  material  per  cubic 

unit  and  the  length  perpen- 
dicular to  the  plane  of  the 
drawing  be  unity.  The  total 
radial  centrifugal  force  due 
to  the  angular  velocity  GO  is 

FIG.  60.  Q  =  

o 

and  the  centrifugal  force  per  square  unit  is 

Q 


which  acts  upon  the  hoop  in  the  same  manner  as  the  internal 
pressure  of  fluid  in  a  pipe  (Art.  9).  Let  6"  be  the  tangential 
tensile  unit-stress  caused  in  the  hoop ;  then  for  equilibrium 

2rp  =  2/5, 
and  hence  the  value  of  5  is 

Tb        w  w 

S=f  =  -^=-(2nrn)\ 

which  shows  that  the  tensile  unit-stress  in  a  thin  hoop  is  r  times 
the  centrifugal  force  of  a  cubic  unit  of  the  revolving  material. 

For  example,  let  it  be  required  to  find  the  unit-stress  in  a 
cast-iron  hoop  2  inches  thick,  4  inches  wide,  and  62  inches 
outer  diameter  when  making  300  revolutions  per  minute. 
Here  w  =  450/1728  pounds  per  cubic  inch,  g  —  32.16  X  12 
inches  per  second  per  second,  r  =  30  inches,  n  =  5  revolutions 
per  second  ;  then  5  is  found  to  be  630  pounds  per  square  inch, 
which  is  a  safe  value  for  cast  iron  in  tension  under  such  con- 
ditions. 

Prob.  147.  A  cast-iron  bar  is  3  X  2  inches  in  section,  and 
9  feet  long.  Through  the  middle  and  normal  to  the  flat  side 
is  a  hole  f  inches  in  diameter.  If  the  bar  be  revolved  around 
an  axis  through  this  hole,  how  many  revolutions  per  second 
will  produce  rupture? 


ART.  IO5.  SHRINKAGE  OF  HOOPS.  233 

ART.  105.    SHRINKAGE  OF  HOOPS.  * 

Hoops  and  tires  are  frequently  turned  with  the  interior 
diameter  slightly  less  than  that  of  the  wheels  or  cylinders  upon 
which  they  are  to  be  placed.  They  are  then  expanded  by- 
heat  and  placed  in  position,  and  upon  cooling  are  held  firmly 
in  position  by  the  radial  stress  thus  produced.  The  effect  of 
this  radial  stress  is  to  cause  tension  in  the  hoop,  and  compres- 
sion throughout  the  mass  that  it  encircles. 

When  the  hoop  is  thin  compared  to  the  diameter  of  the 
cylinder  upon  which  it  is  to  be  shrunk,  the  entire  stress  due  to 
the  shrinkage  may  be  practically  regarded  as  confined  to  the 
hoop.  The  tangential  unit-stress  in  the  hoop  will  then  be  due 
only  to  the  increase  in  length  of  the  circumference,  and  this 
will  be  proportional  to  the  increase  in  its  diameter. 

Let  D  be  the  diameter  of  the  cylinder  upon  which  the  hoop 
is  to  be  shrunk,  and  d  be  the  interior  diameter  to  which  the 
hoop  is  turned.  Supposing  that  D  is  unchanged  by  the  shrink- 
age,  d  will  be  increased  to  Dy  and  the  relative  change  in  length 
or  the  unit-elongation  of  the  hoop  will  be 

D-d 
~~d~' 

and  hence  the  unit-stress  produced  will  be 


where  E  is  the  coefficient  of  elasticity  of  the  material. 

A  very  common  rule  for  the  case  of  steel  hoop  shrinkage  is 


to  make  D  •—  d  equal  to  yVfrrA  that  is,  the  hoop  is  turned  so 
that  the  interior  diameter  is  T^ryth  less  than  the  diameter 
of  the  cylinder.  Then 

D-d         i 


or  s  may  be  taken  also  as  y^.     Thus  the  tangential  unit- 


234  TENSION  AND  COMPRESSION.  CH.  X. 

stress  S  will  be  20  ooo  pounds  per  square  inch.  For  wrought 
iron  s  may  be  taken  as  2faQ  and  S  will  be  12  500  pounds  per 
square  inch. 

These  values  of  5  are  too  high,  because  a  part  of  the  effect 
of  shrinkage  is  expended  in  producing  compression  in  the 
cylinder.  In  Art.  142  the  subject  will  be  further  discussed, 
and  the  final  decrease  in  D  as  well  as  the  increase  in  d  will  be 
determined. 

Prob.  148.  Upon  a  cylinder  18  inches  in  diameter  a  wrought- 
iron  hoop  2  inches  thick  is  to  be  placed.  -The  hoop  is  turned 
to  an  interior  diameter  of  17.98  inches  and  shrunk  on.  Com- 
pute the  tensile  unit-stress  in  the  hoop. 

ART.  1 06.    SPHERICAL  ROLLERS. 

Let  a  sphere  of  radius  r  be  placed  between  two  plates  of 
equal  thickness  r,  and  be  subject  to  compression  by  a  load  W. 

The  vertical  diameter  DD  is 
thus  shortened  to  BB,  while 
any  vertical  line  dd  is  short- 
ened to  bb.  Let  the  total 
shortenings  CD  be  called  A, 
and  the  shortenings  cd  at 
any  point  be  called  y.  Let 
the  unit-stresses  at  B  and  b 
be  denoted  by  5"  and  Sy. 
Then,  if  the  elastic  limit  is 
FIG.  61.  not  exceeded, 

5        X  S 

s,=y  or   ^  =  i* 

The  unit-stress  5  is  evidently  the  maximum,  and  it  is  required 
to  express  it  in  terms  of  W,  r,  and  rr  This  will  now  be  done 
approximately,  assuming  that  each  vertical  element  acts  inde- 
pendently of  the  others  and  that  no  lateral  bulging  of  the 
sphere  occurs. 


ART.  I06.  SPHERICAL  ROLLERS.  235 

The  value  of  A  may  be  expressed  by  noting  that  BD  is  the 
shortening  of  the  radius  r,  and  CB  that  of  the  thickness  rl  ; 
thus 


in  which  E  and  El  are  the  coefficients  of  elasticity  of  the 
sphere  and  plates  respectively.  Also  the  sum  of  all  the  verti- 
cal stresses  in  the  spherical  segment  PDF  must  equal  the  total 
load,  or 

=         ydA, 


where  dA  denotes  an  elementary  area  of  the  circle  whose 
radius  is  CF.  Thus  two  equations  have  been  found  connect- 
ing the  two  unknown  quantities  5  and  A. 

To  solve  these  equations  consider  that  y  is  any  ordinate  cd 
of  the  spherical  segment  corresponding  to  an  abscissa  Cc  or  x. 

Thus  fydA  is  the  volume  of  the  segment,  which  is  very  nearly 

equal  to  one-half  the  cylinder  having  CF  as  the  radius  of  its 
base  and  CD  as  its  altitude,  since  the  arc  of  contact  FBF  is 
very  small.  Now  CD  =  A  and  CF  =  i/2r\  —  A'  = 
nearly,  and  hence 

W  =  T  •  -  •  7T2r\  = 

A.     2 

Inserting  for  \  its  value,  this  becomes 


which  is  an  approximate  formula  for  the  investigation  of 
spherical  rollers  when  the  upper  and  lower  plates  are  of  equal 
thickness. 

Compression  tests  made  upon  spheres  between  plane  sur- 
faces show  that  usually  the  plates  are  not  indented  as  shown 
by  FBFC  in  Fig.  61,  but  that  the  sphere  alone  suffers  mate- 


236  TENSION    AND    COMPRESSION,  CH.  X. 

rial  deformation.  In  this  case  BC  is  zero,  and  thus  the  last 
formula  applies  independently  of  the  thickness  of  the  plates 
if  rl/El  be  made  zero.  The  practical  formula  for  spherical 
rollers  of  radius  r  then  is 

W  =  nS*r*/E, 

or,  the  strength  of  a  sphere  varies  with  the  square  of  its 
radius.  Strictly  this  formula  only  applies  when  the  elastic 
limit  is  not  exceeded,  but  for  cases  of  rupture  W=  Cr*  is  an 
empirical  formula,  C  being  a  constant  found  by  experiment 
for  each  kind  of  material. 

Prob.  149.  How  many  steel  spherical  rollers  are  required 
for  a  load  of  6  ooo  pounds  and  a  working  stress  of  1  5  ooo 
pounds  per  square  inch  if  r  =  2  inches,  and  also  if  r  =  6 
inches? 

ART.  107.     CYLINDRICAL  ROLLERS. 

The  reasoning  of  the  last  article  applies  also  to  a  cylindrical 
roller  included  between  two  plates  of  equal  thickness.  Let 
Fig.  6  1  represent  a  cross-section  perpendicular  to  the  axis  of 
the  roller  whose  length  is  /.  Then,  as  before, 


are  the  two  equations  for  determining  5.  The  area  A  is  that 
of  a  plane  with  width  FF  and  length  /,  so  that  jydA  is  the 

volume  of  the  cylindrical  segment  whose  cross-section  is  FDF. 
Since  the  area  of  FDF  is  very  small,  it  is  closely  two-thirds  of 
the  rectangle  whose  base  is  FF  and  altitude  is  CD.  Now  CD 
=  A  and  FF  =  2  ^2r\  very  nearly,  and  hence 


.  2 


ART.   ID/.  CYLINDRICAL   ROLLERS.  237 

and  this  becomes,  after  inserting  the  value  of  \ , 


which    is  an   approximate    formula  for   the   investigation   of 
cylindrical  rollers. 

In  making  tests  on  rollers  it  is  found  that  the  plates  are  usu- 
ally but  little  indented,  most  of  the  deformation  being  in  the 
roller.  The  above  investigation  may  be  adapted  to  this  case  by 
making  rl/£1  equal  to  zero,  and  the  last  formula  then  becomes 


•=M/— 

V    E 


which  shows  that  the  strength  of  a  cylindrical   roller  varies 
directly  as  its  radius.    If  w  be  the  load  per  unit  of  length,  then 

'Is5 


Taking  5=  15000  and  E  =  30000000  pounds  per  square 
inch  for  steel  or  cold  rolled  iron,  this  reduces  to  w  =  6$or  or 
w  =  315^,  where  d  is  the  diameter  of  the  roller. 

It  should  be  noted  that  a  common  rule  used  in  proportion- 
ing cylindrical  rollers  for  bridge  seats  is  to  take  the  safe  load 
in  pounds  per  linear  inch  as  500  V d,  thus  making  the  strength 
vary  with  the  square  root  of  the  diameter.  This  practice 
seems  to  be  based  upon  the  authority  of  GRASHOF,  who  was 
the  fisst  to  deduce  the  formula  at  the  top  of  this  page,  but 
who  appears  to  have  placed  a  doubtful  interpretation  upon  it. 
In  general,  the  true  rule  is  that  the  load  should  be  nearly 
proportional  to  the  diameter  of  the  roller. 

Prob.  150.  A  load  of  192  ooo  pounds  is  carried  on  wrought- 
iron  rollers  16  inches  long  and  3  inches  in  diameter.  How 
many  rollers  are  required  if  5  is  to  be  12  ooo  pounds  per 

square  inch  ? 


238 


TENSION    AND  COMPRESSION. 


CH.  X. 


ART.  108.     ECCENTRIC  LOADS. 

Let  a  load  P  be  applied  to  the  end  of  a  short  bar  at  a  hori- 
zontal distance  /  from  the  center  of  gravity  of  its  cross-section, 
as  shown  in  Fig.  62.  If  two  forces  equal  to  P 
and  acting  in  opposite  directions  be  supposed 
to  be  applied  to  the  end  of  the  bar,  the  equi- 
librium is  undisturbed,  and  it  is  seen  that  the 
downward  force  produces  pure  tension,  while 
the  couple  formed  by  P  and  the  other  force 
produces  flexure.  Let  A  be  the  cross-section 
of  the  bar  and  >S  the  maximum  unit-stress 
caused  by  the  flexure ;  then  the  resultant  unit- 
stresses  are 

P  _P 

1  ~~  A  '      '         *  ~~  A  '"    ' 
the  former  being  on  the  side  of  the  bar  nearest 
FIG.  62.  to  p  ancj  tne  latter  on  the  other  side. 

Let  /  be  the  moment  of  inertia  of  the  cross-section  with 
respect  to  an  axis  through  the  center  of  gravity  and  normal  to 
the  direction  of  the  arm  /.  Let  r  be  the  radius  of  gyration  of 
the  cross-section  so  that./  =  Ar*.  Let  cl  and  ct  be  the  dis- 
tances from  the  axis  to  the  sides  of  the  section  nearest  to  and 
farthest  from  P.  Then  from  the  formula  (4)  of  Art.  21  the 
two  values  of  S  are  found,  and  the  above  expressions  reduce 
to  the  practical  formulas 


which   give  the  greatest  and  least  tensile  stresses  under  the 
eccentric  application  of  the  load. 

These  formulas  apply  also  to  compression  under  an  eccen- 
tric load,  provided  that  the  bar  be  short  so  that  no  lateral 
flexure  can  occur.  As  an  example  let  a  short  block  have  a 


ART.  1 08. 


ECCENTRIC   LOADS. 


239 


rectangular  section,  the  dimension  parallel  to  the  arm  p  being 
d,  while  the  one  normal  to  it  is  b.     Then  the  formulas  reduce 


and  in  Fig.  63  are  shown  the  distribution  of  stresses  for  several 
values  of  /.  In  the  first  sketch  P  is  applied  at  the  center  of 
the  section  so  that  the  unit-stresses  are  uniform  and  both  S 


and  S,  are  equal  to  -j.     In  the  second  sketch/  is  taken  as 


p 
c 

p 
c 

p 

c 

c  /flll 

\ 

I 

1 

m 

JIP1* 

£ 

' 

p 

JP* 

Si 

( 

f 

f 

FIG.  63. 

3  P  l  P 

which  gives  £,=  ---  and  S,  =  -  — .     In  the  third  sketch  /  is 

2  si  2  A 

p 

taken  as  \d,  so  that  St  =  2—  and  S9  ^=  o.     As  the  load  moves 

A 

further  away  from  the  center  C,  the  stress  5",  increases,  while 
St  becomes  negative,  showing  that  it  is  tension.  Thus  in  the 
last  sketch  where  p  is  equal  to  \d  the  compressive  unit-stress 

P  P 

5,  is  3  -r,  while  the  tensile  unit-stress  5,  is  2  -j.     In  all  these 

./i  A 

P 

cases  the  unit-stress  at  the  center  C  is  the  mean  value  -r. 

It  is  thus  seen  that  the  eccentric  application  of  a  load  may 
materially  increase  the  direct  stress  of  tension  or  compression. 
In  the  case  of  a  rectangular  masonry  pier  the  greatest  devia- 
tion of  P  from  the  center  should  never  be  greater  than  \d,  in 


240  TENSION  AND   COMPRESSION.  CH.  X. 

order  that  no  tension  may  be  brought  upon  the  joint.  In 
other  words  the  point  of  application  of  P  should  be  kept 
within  the  *  middle  third  '  of  the  base. 

In  both  the  cases  above  discussed  the  body  under  tension  or 
compression  has  been  supposed  to  be  very  short,  so  that  no 
material  lateral  movement  can  result.  If  in  Fig.  62  the  bar  be 
long  the  arm  /  will  be  different  for  different  parts  of  the 
length,  and  the  greatest  bending  moment  will  occur  at  the 
lower  end,  so  that  the  formulas  given  apply  only  to  that  end, 
while  for  other  sections  they  give  results  too  large.  For 
compression,  however,  the  reverse  is  the  case,  and  application 
can  only  be  safely  made  to  blocks  whose  length  does  not 
exceed  ten  times  the  thickness  (see  Art.  62). 

Prob.  151.  A  bar  of  circular  cross-section,  2  inches  in  diam- 
eter, is  under  tension  by  a  load  of  12000  pounds  which  is 
applied  at  0.5  inches  from  the  center.  Compute  the  maximum 
tensile  unit-stress. 


ART.  109.  THE  WORK  OF  FLEXURE.  241 


CHAPTER    XL 
FLEXURE  OF   BEAMS. 

ART.  109.    THE  WORK  OF  FLEXURE. 

This  subject  was  treated  in  Art.  96,  but  it  will  now  be  dis- 
cussed again  in  order  to  deduce  a  more  general  relation  be- 
tween the  work  of  the  external  forces  and  that  of  the  internal 
stresses. 

Let  P  be  a  load  upon  a  beam  and  A  the  deflection  under  it. 
The  load  being  gradually  applied  the  work  done  by  P  is  %Pd9 
and  this  must  be  equal  to  the  internal  work  or  resilience  of 
the  molecular  stresses. 

For  a  beam  loaded  uniformly  with  w  per  linear  unit  let  y 
be  the  deflection  at  any  point  whose  abscissa  is  x.  Then  the 
load  wdx  on  the  short  distance  dx  deflects  the  amount  j,  and 
the  elementary  external  work  is  \wydx.  Thus  the  integral  of 
this,  if  y  is  known  in  terms  of  x,  will  give  the  total  work  of 
the  uniform  load,  if  the  integration^  be  extended  over  the 
entire  length  of  the  beam.  This  will  be  equal  to  the  internal 
work,  or  resilience. 

A  general  expression  for  the  resilience,  or  internal  work  of 
the  horizontal  fibers,  in  terms  of  the  bending  moment  M  of 
the  applied  forces  will  now  be  deduced  ;  it  is  applicable  to  all 
cases  in  which  the  elastic  limit  of  the  material  is  not  surpassed 
by  the  maximum  fiber  stress  S. 

When  a  beam  deflects  under  the  action  of  a  load  the  hori- 
zontal fibers  upon  one  side  of  the  neutral  surface  are  elongated 
and  upon  the  other  side  are  compressed.  The  internal  work 
will  be  found  by  taking  the  sum  of  the  products  formed  by 


242  FLEXURE   OF  BEAMS.  CH.  XI. 

multiplying  the  stress  upon  any  elementary  area  by  its  elonga- 
tion or  compression. 

Using  the  same  notation  as  in  Chapter  III.,  the  horizontal 
unit-stress  at  any  distance  z  from  the  neutral  axis  is  represented 

by  —  .     In  the  distance  dx  the  elongation  or  compression  due 

C*       ^7 

to  this  unit-stress,  is  by  (2)  found  to  be  —  —  .    The  elementary 

work  of  a  fiber  of  the  area  a  under  this  gradually  applied  unit- 
stress  hence  is, 

i    Saz        Szdx 
~2'~7~          cE  * 

The  work  done  in  the  distance  dx  by  all  the  fibers  in  the  cross- 

section  now  is, 

,„      S'Zaz*  , 
dK  =  -  irr-dx. 

2?E 

S*  .   M* 
Here  ^a£  =  /and  from  formula  (4),  the  value  of  —5-  is  -— 

TM       f  77^-  * 

Therefore  dK  = 


. 

2EI 

This  is  the  formula  for  the  work  done  in  the  distance  dx.  By 
expressing  M  as  a  function  of  x,  and  integrating,  the  total  in- 
ternal work  K  between  assigned  limits  can  be  found. 

For  example,  consider  a  cantilever  beam  loaded  at  the  end 
with  a  weight  P.  Here  M  =  —  Px.  Inserting  this  and  in- 
tegrating between  the  limits  o  and  /,  gives, 


for  the  total  internal  work  in  the  beam  due  to  a  load  which  is 
gradually  applied. 


ART.  109.  THE  WORK  OF   FLEXURE.  243 

The  preceding  furnishes  a  new  method  of  deducing  the  de- 
flection of  a  beam  loaded  with  a  single  weight  P.  Let  A  be 
the  deflection  under  the  weight.  Then  \PA  is  the  external 
work  done  by  the  load  P  upon  the  beam,  and  this  must  equal 
the  internal  work  K.  Hence  the  formula, 

PA-     CM*dx 

from  which  A  may  be  found  for  particular  cases. 

For  example,  consider  a  cantilever  beam  loaded  at  the  end 
with  P.     Then  the  internal  work  is,  as  shown  above, 
Hence  the  deflection  A  is, 

A  =-.  Pl* 


which  is  the  same  as  otherwise  found  in  Art.  34. 

Px 

For  a  simple  beam  loaded  at  the  middle  the  value  of  Mis  — 

2 

and  then 


from  which  the  deflection  is, 


which  is  the  same  as  found  in  Art.  35  by  the  use  of  the  elastic 
curve. 

Prob.  152.  Prove  that  the  internal  work  caused  by  a  uni- 
formly distributed  load  on  a  cantilever  beam  is  ^ths  of  that 
caused  by  the  same  load  applied  at  the  end. 

Prob.  153.  Deduce  by  the  method  of  Art.  35,  and  also  by 
the  use  of  the  principle  of  internal  work,  the  deflection  under 
a  load  P  which  is  placed  upon  a  simple  beam  at  a  distance  £/ 
from  one  end. 


244  FLEXURE  OF  BEAMS.  CH.  XI. 


ART.  1 10.    STATIC  AND  SUDDEN  DEFLECTIONS. 

A  static  deflection  is  one  produced  by  the  gradual  applica- 
tion of  the  load,  so  that  at  each  instant  the  beam  is  in  a  condi- 
tion of  static  equilibrium  and  hence  no  oscillations  occur.  The 
formulas  for  the  deflection  of  beams  deduced  in  Chapters  III 
and  IV,  as  well  as  the  discussion  of  the  last  Article,  are  all  static 
deflections  where  the  elastic  limit  of  the  material  is  not  sur- 
passed. In  such  case  the  load  increases  from  o  up  to  its  maxi- 
mum value  P,  and  simultaneously  the  deflection  increases 
from  o  up  to  its  maximum  value  A.  If  Q  be  the  load  at  any 
instant  and  8  the  corresponding  deflection,  then  the  deflections 
are  proportional  to  the  loads  or  6/A  =  Q/P. 

A  load  is  said  to  be  suddenly  applied  when  it  acts  with  uni- 
form intensity  during  the  full  period.  For  instance,  let  a  load 
be  attached  to  a  ring  which  is  placed  around  the  middle  of  a 
beam,  and  let  the  load  be  supported  so  .that  the  ring  just 
touches  the  upper  surface  of  the  beam ;  then  if  the  support  of 
the  load  be  suddenly  withdrawn  the  force  of  gravity  acts  upon 
the  load  with  uniform  intensity  during  the  entire  period  of  de- 
flection. In  this  case  the  maximum  deflection  is  greater  than 
for  a  static  load,  but  as  soon  as  it  is  reached  a  series  of  oscilla- 
tions occurs  until  finally  the  beam  comes  to  rest  with  a  deflec- 
tion due  to  the  static  load.  The  maximum  stress  in  the  beam 
evidently  occurs  at  the  instant  of  greatest  deflection.  If  S  be 
the  unit-stress  due  to  the  static  load  P  under  the  deflection  A, 
and  if  T  be  the  unit-stress  due  to  the  sudden  load  P  under  the 
deflection  tf,  both  loads  being  applied  at  the  same  point  on  the 
same  beam,  then  from  Art.  37  the  stresses  are  proportional  to 
the  deflections,  or  6/A  =  775. 

The  deflection  under  a  sudden  load  is  double  that  under  the 
same  static  load,  and  the  stress  under  a  sudden  load  is  double 
that  under  the  same  static  load.  In  order  to  prove  this  refer 


ART.   1  10.        STATIC   AND   SUDDEN   DEFLECTIONS.  245 

to  Art.  96  and  note  that  the  total  internal  work  or  resilience 
due  to  the  maximum  unit-stress  S  in  the  beam  is 

K=\?E.C.Al, 

where  Al  is  the  volume  of  the  beam,  and  C  is  a  constant  de- 
pending upon  the  shape  of  its  cross-section  and  the  arrange- 
ment of  its  ends.  Thus  for  a  given  beam  the  internal  work 
varies  as  S2.  Now  for  the  gradually  applied  load  P  the  deflec- 
tion is  A  and  the  external  work  is  \PA.  Hence  as  internal 
work  equals  external  work, 


which  gives  the  relation  between  A  and  5.  Again  for  the 
sudden  load  the  deflection  is  d  and  the  maximum  unit-stress  is 
T\  the  external  work,  however,  is  P3,  since  the  load  acts  with 
full  intensity  during  the  entire  period  of  deflection,  while  the 
stress  increases  gradually  from  o  up  to  T\  hence 

i  rf 
PS  =  -  -jCAl, 

which  gives  the  relation  between  d  and  T.  By  comparing 
these  two  equations  there  is  found 


and  remembering  that  T/S  =  <?/^,  as  shown  above,  this  gives 

6  =  24,     and     T  —  25, 
which  proves  the  proposition  as  stated. 

Another  method  of  establishing  the  same  truth  is  to  com- 
pare the  sudden  load  P  with  the  static  load  W  which  will  pro- 
duce the  same  deflection  A  and  hence  the  same  unit-stress  5. 
The  internal  work  is  thus  the  same  in  the  two  cases,  but  the 
external  work  of  the  sudden  load  is  PA,  while  that  of  the  static 
load  is  %WA.  Therefore  P  =  %W\  that  is,  the  sudden  load 
that  produces  a  given  stress  is  one-half  of  the  static  load  that 
produces  the  same  stress. 


246  FLEXURE   OF  BEAMS.  CH.  XL 

These  propositions  are  true  only  when  the  maximum  stresses 
caused  by  the  deflections  are  within  the  elastic  limit  of  the 
material,  since  all  the  reasoning  by  which  they  are  established 
supposes  the  law  of  proportionality  of  stress  to  deformation  to 
be  observed.  It  is  hence  not  to  be  expected  that  they  can  be 
verified  by  the  rupture  of  beams.  It  has,  however,  been  found 
to  be  approximately  true  for  cast  iron,  but  for  wrought  iron 
and  steel  the  sudden  load  that  produces  rupture  is  greater 
than  one-half  of  the  static  load. 

The  following  experiments  of  HODGKINSON  on  cast-iron 
beams  illustrate  very  well  the  agreement  of  theory  with  prac- 
tice near  the  elastic  limit  and  the  disagreement  when  the 
elastic  limit  is  exceeded.  Each  beam  was  laid  on  supports  9. 
feet  apart,  and  a  lever  at  the  middle  prevented  deflection  when 
the  load  was  applied.  This  was  hung  to  the  beam  by  two 
rings,  one  on  each  side  of  the  lever  and  placed  as  close  to  it  as 
possible  without  touching.  The  lever  being  instantaneously 
removed  the  load  was  brought  suddenly  into  action,  and  the 
deflection  was  registered  upon  a  vertical  sheet  of  paper  by 
means  of  a  pencil  screwed  to  the  side  of  the  beam.  Before 
applying  the  loads  in  this  sudden  manner  the  beams  were  tested 
in  the  usual  way  by  gradually  applied  loads.  The  results  here 
given  are  the  mean  of  two  or  three  tests  upon  different  beams. 
It  will  be  found  that  for  each  beam  the  first  load  gives  a  unit- 
stress  less  than  the  elastic  limit,  while  the  second  gives  a 


Size  of  Beam. 
Inches. 

Load. 
Pounds. 

Deflection 
Gradual. 

in  Inches.    Ratio  of  Deflections, 
Sudden.  Sudden  to  Gradual. 

I   X  2 

112 

0.253 

0.515 

2.04 

I    X  2 

224 

O.58O 

0-933 

1.61 

IX  3 

224 

0.163 

0.303 

1.86 

I  X  3 

560 

0.4IO 

0.720 

1.76 

4X  it 

448 

0.770 

1.510 

1.96 

4  X  it 

784 

1.275 

2.225 

i-73 

greater  value.     Thus  for  the  first  beam  under  112  pounds  at 


ART.  III.  DEFLECTION   UNDER   IMPACT.  247 

the  middle  S  is  4  540,  and  for  224  pounds  5  is  9070  pounds 
per  square  inch. 

The  breaking  loads  of  these  beams  were  also  observed  under 
both  gradual  and  sudden  loads  with  the  following  results : 

Size  of  Beam.  Breaking  Load  in  Pounds.         Ratio  of  Loads, 

Inches.  Gradual.  Sudden.      Gradual  to  Sudden, 

1X2  IOOO  569  1.76 

i  X  3  2008  1219  1.65 

4X4  1911  1082  1.77 

Here  the  ratio  is  seen  to  be  about  the  same  as  for  the  heavier 

loads  in  the  cases  of  deflection. 

KIRKALDY'S  experiments  on  wrought-iron  beams  show 
that  the  ratio  of  the  gradual  to  the  sudden  breaking  load 
ranges  between  1.2  and  1.3,  so  that  the  elastic  law  does  not 
even  approximately  apply.  There  is  a  very  good  reason  why 
it  should  not  apply,  and  for  this  the  student  may  consult  the 
Proceedings  of  the  Engineers'  Society  of  Western  Pennsyl- 
vania for  May,  1894. 

Prob.  154.  Compute  the  static  deflection  and  the  unit-stress 
5  for  the  third  experiment  on  the  cast-iron  beams  given  above. 

ART.  in.    DEFLECTION  UNDER  IMPACT. 

As  the  deflection  under  a  sudden  load  is  greater  than  for  a 
static  load,  so  the  deflection  under  impact  is  greater  still.  A 
load  falling  through  a  small  height  upon  a  beam  may  easily 
produce  a  deflection  and  a  stress  that  may  prove  dangerous  to 
its  stability.  In  Arts.  93  and  103  rules  were  deduced  for  bars 
subject  to  impact,  and  a  similar  investigation  leading  to  similar 
results  will  now  be  made  for  beams. 

Let  the  beam  be  placed  in  a  horizontal  position  and  let  a 
weight  P  move  horizontally  with  the  velocity  v  so  as  to  strike 
it  at  the  middle  and  cause  a  lateral  deflection.  The  work 


248  FLEXURE  OF   BEAMS.  CH.  XI. 

V* 

stored   up   by  the  moving  weight  is  P—9  or  Ph  if  h  be  the 

o 

height  due  to  the  velocity.  Let  W  be  the  weight  of  the  beam, 
6  the  maximum  deflection  produced  by  the  impact,  and  T  the 
corresponding  maximum  unit-stress.  Let  A  be  the  deflection 
due  to  a  static  load  Pt  and  S  be  the  corresponding  maximum 
unit-stress.  Then  d/A  =  T/S,  or  the  deflections  are  propor- 
tional to  the  stresses.  Also,  let  Q  be  a  static  load  which  will 
produce  the  deflection  d,  then  likewise  Q/P  =  <$/A,  or  the 
static  loads  are  proportional  to  the  deflections.  The  internal 
work  or  resilience  of  the  beam  now  is  %Qd,  and  this  must  equal 
the  external  work  Ph,  or 

±Q6  =  ±.2f.d  =  Ph. 

Also  substituting  for  tf  its  value  in  terms  of  T,  S,  and  A,  an 
equation  between  5  and  T  results.  Thus, 

(22)         * 

are  the  formulas  for  finding  the  maximum  deflection  and  cor- 
responding unit-stress  in  a  simple  beam  under  impact  at  the 
middle,  A  and  5  being  given  by  the  expressions  of  Chapter  III, 


~  mEP  ~  n' 

where  /  is  the  length  of  the  beam,  m  and  n  numbers  depending 
upon  the  arrangement  of  its  ends,  /  the  moment  of  inertia  of 
the  cross-section,  and  c  the  distance  from  the  neutral  axis  to 
the  fiber  whose  stress  is  5  or  T. 

The  above  results  will  be  somewhat  modified  by  the  resist- 
ance of  inertia  of  the  beam,  as  in  the  case  of  longitudinal  im- 
pact discussed  in  Art.  103.  An  instant  before  the  contact  the 
weight  P  has  the  velocity  v  and  the  beam  is  at  rest  ;  when  the 
contact  is  complete  both  P  and  the  middle  of  the  beam  are 
moving  with  a  common  velocity  V.  Plainly  V  is  less  than  v, 


ART.  III.  DEFLECTION   UNDER   IMPACT.  249 

V* 

as  a  portion  of  the  energy  P  —   has  been  communicated  to  the 

beam  whose  weight  is  W.  When  the  velocity  V  is  acquired 
any  element  dW  is  moving  with  the  velocity  u,  and  the  com- 
bined energy  of  weight  and  beam  then  is 

V*  f*         u* 

K  =  P--  dW—  = 


V*  f* 

=  P--  +     / 

*g  r  J 


and  this  is  the  energy  which  is  effective  in  producing  the  de- 
flection. But  also,  as  soon  as  the  velocity  V  is  acquired  the 
condition  of  impact  of  must  obtain,  namely, 

Pv  =  (P+qW)V. 

Eliminating  V  from  these  equations  gives 
_         />V  Ph 

~~~-~-  i  +  qk> 


in  which  k  is  the  ratio  W/P  and  q  is  a  number  whose  value  is 
to  be  determined. 

The  above  value  of  K  being  placed  equal  to  J<2#,  as  before, 
there  are  found 

(22y      *  =  X/^g,        T=S^^},       , 

as  the  modified  formulas  for  deflection  and  stress  due  to  hori- 
zontal impact. 

It  now  remains  to  determine  the  value  of  ^,  and  this  will  de- 
pend upon  the  arrangement  of  the  ends  of  the  beam.  The 
general  value  of  q  is  seen  to  be 

and  the  integration  is  to  be  extended  over  the  entire  length  of 
the  beam.  Now  dW/W  =  dx/l\  also  if  y  be  the  deflection  at 
any  point  and  y^  the  deflection  at  the  point  whose  velocity  is 
F,  then  u/  V  =  y/yr  Thus  the  integral  becomes 


/>vf 
= J  y;i> 


FLEXURE   OF   BEAMS.  CH.  XL 

which  may  be  applied  to  any  particular  case  where  y/y^  can  be 
expressed  as  a  function  of  x. 

For  a  beam  supported  at  the  ends  and  loaded  in  the  middle 
the  ordinate  y  of  the  elastic  curve  in  terms  of  the  maximum 
deflection^  and  the  abscissa  x  is,  from  Art.  35, 


and  for  this  case  the  integral  becomes 


Therefore  for  a  beam  supported  at  the  ends  and  impinged 
upon  in  the  middle  there  obtains 


which  is  the  value  to  use  in  formula  (22)'.  This  result  was 
first  established  theoretically  by  Cox  in  1848,  but  HODGKIN- 
SON  had  previously  found  by  experiments  on  cast-iron  beams 
that  the  value  of  q  was  about  \. 

As  a  numerical  example  let  a  cast-iron  beam  on  supports  9 
feet  apart  be  I  inch  wide  and  2  inches  deep,  and  have  a  static 
load  of  50  pounds  at  the  middle.  The  deflection  and  stress  at 
the  middle  due  to  this  load  are 

A  —  0.131  inches,     5  =  2025  pounds  per  square  inch. 
Now  suppose  that  this  load  of   50  pounds  moves  horizontally 
with  a  velocity  due  to  a  fall  of  £  inch.     Then  from  (22) 

d  =  0.362  inches,     T=  5590  pounds  per  square  inch, 
which  are  the  results,  supposing  that  the  beam  has  no  weight. 
Taking  this  into  account,  the  weight  W  is  56.4  pounds,  whence 
'£  =  1.128,  and  by  (22)' 

d  =  0.290  inches,     T  =  4470  pounds  per  square  inch, 
which  are  more  exact  results. 

It  is  seen  by  this  investigation  that  very  small  velocities  of 


ART.  III.  DEFLECTION    UNDER   IMPACT.  2$  I 

impact  may  produce  very  high  stresses  in  a  beam.  Thus  in 
(22)  the  static  deflection  A  is  always  small,  and  if  h  =  2-J,  Tis 
28.  It  is  also  seen  that  the  influence  of  the  resisting  inertia  of 
the  beam  increases  with  k,  that  is,  with  the  ratio  of  the  weight 
of  the  beam  to  the  impinging  weight. 

When  the  weight  falls  vertically  upon  the  horizontal  beam, 
h  being  the  height  of  fall  to  the  top  of  the  beam,  the  formulas 
(22)  and  (22)'  are  to  be  modified  slightly,  since  the  external 
work  is  P(h  -f-  6).  Thus  are  found 


which  are  seen  to  be  the  same  in  form  as  those  deduced  for 
longitudinal  impact  in  Art.  103.  For  instance,  if  the  load  in 
the  above  example  drops  %  inch,  then  $  =  0.295  inches  instead 
of  0.290  inches. 

The  interesting  experiments  made  by  KEEP  in  1899  enable 
a  comparison  of  theory  and  practice  to  be  made.  A  hori- 
zontal bar  of  cast  iron  I  X  i  X  24  inches  was  loaded  with 
weights  of  25,  50,  75,  and  100  pounds,  and  the  corresponding 
static  deflections  were  found  to  be  0.0448,  0.0896,  0.1344, 
and  o.  1792  inches.  They  were  then  struck  laterally  by  ham- 
mers of  the  same  weights  which  swung  like  pendulums  and 
had  a  vertical  fall  of  2  inches.  The  dynamic  deflections  due 
to  these  weights,  as  carefully  measured  by  a  graphic  recording 
apparatus,  are  given  below.  From  the  given  data  the  theo- 
retic deflections  under  impact  have  been  computed  from 
formula  (22)',  and  the  following  is  a  comparison  of  observed 
and  theoretic  values  : 

Swinging  P  =  25  50  75  100    pounds 

Observed  3   =         0.122         0.150        0.175         0.200  inches 
Theoretic  d    =         o.ioo         0.145         0.180         0.209  inches 


FLEXURE   OF   BEAMS.  CH.  XI. 

Experiments  were  also  made  by  allowing  the  same  weights  to 
fall  vertically  on  the  bar  through  heights  of  2  inches.  For- 
mula (22)"  applies  to  this  case,  and  the  following  is  a  com- 
parison of  the  dynamic  deflections  as  observed  and  computed  : 

Falling        P  =  25  50  75  100    pounds 

Observed    8  =         0.130         0.159         0.181         0.209  inches 
Theoretic    S  =         0.103         0.152         0.188         0.220  inches 

It  is  seen  that  the  comparison  is  very  satisfactory ;    it  would 
be  expected,    however,    that  the  observed   values  should   be 
always  slightly  less  than  the  theoretic  ones,  because  some  of 
the  work  of  impact  is  probably  expended  in  producing  heat. 
Prob.  155.   Check  some  of  the  above  values  of  theoretic  <?. 

ART.  i  n£.     OSCILLATIONS  AFTER  IMPACT. 

The  formulas  of  the  last  article  apply  to  the  momentary 
maximum  deflection  under  the  impact  of  the  weight.  A 
series  of  oscillations  then  ensues,  and  finally,  if  the  weight 
remains  on  the  beam,  it  comes  to  rest  with  a  deflection  due 
to  the  static  load.  The  following  figure,  illustrating  these 
phenomena  in  an  interesting  manner,  is  from  an  automatic 
record  taken  by  P.  H.  DUDLEY  in  1895.  A  railroad  rail  30 
feet  long  and  weighing  80  pounds  per  yard  was  placed  at  its 
extreme  ends  upon  rigid  supports  and  had  a  scale-pan  of  210 
pounds  weight  suspended  from  the  middle.  Secured  to  the 
center  of  the  rail  was  an  attachment  carrying  a  pencil  which 
recorded  upon  paper  moving  horizontally  the  deflections  and 
oscillations  of  the  rail.  A  load  of  100  pounds  being  suddenly 
applied,  the  maximum  deflection  was  0.240  inches,  but  when 
the  beam  came  to  rest  the  static  deflection  was  found  to  be 
o.  120  inches.  The  weight  of  100  pounds  was  next  dropped 
from  the  height  of  12  inches,  and  the  maximum  deflection  was 
0.910  inches;  about  240  oscillations  then  ensued,  as  shown  in 
the  figure,  and  in  about  42  seconds  the  rail  came  to  rest. 


ART.  112. 


PRESSURE    DUE   TO    IMPACT. 


253 


By  an  investigation  exactly  similar  to  that  of  Art.  103 
it  may  be  shown  that  the  time  of  one  oscillation  of  a  simple 
beam,  when  struck  either  horizontally 
or  vertically,  is  given  by 


g 

in  which  k  is  the  ratio  of  the  weight 
of  the  beam  to  that  of  the  falling 
load,  or  k  =  W/P.  If  P  falls  in  a 
scale-pan  whose  weight  is  Wlt  then 
I  _|_  \^k  is  to  be  replaced  by  I  +  £,  + 
\\k,  in  which  k,  is  the  ratio  WJP. 
For  the  above  case  P=  100,  Wl=2\o, 
and  W  —  800  pounds  ;  also  A  =.  o.  120 
inches;  and  for  these  data  the  formula 
gives  /  =  0.146  seconds  for  the  time 
of  one  oscillation,  and  accordingly  the 
theoretic  time  for  240  oscillations  is 
about  35  seconds;  owing  to  air  resis- 
tance, however,  this  was  prolonged  to 
about  42  seconds. 

Prob.  156.  Prove  that  ||  is  to  be 
replaced  by  ^|  for  the  case  of  impact 
against  a  beam  fixed  at  both  ends. 

ART.  112.     PRESSURE  DUE  TO 
IMPACT. 

When  a  weight  P  falls  from  a 
height  h  upon  a  beam,  a  pressure  is 
produced  at  the  point  of  contact.  This 
pressure  is  variable  during  the  period 
of  deflection,  being  zero  at  the  first 
instant  of  contact,  rapidly  increasing  to  a  maximum,  and  then 
decreasing,  until  at  the  moment  of  greatest  deflection  there 


254  FLEXURE  OF  BEAMS.  CH.  XI. 

exists  a  pressure  sufficient  to  cause  the  weight  to  rebound  a 
short  distance.  The  laws  governing  the  variation  of  the  press- 
ure are  not  understood,  but  the  mean  or  average  pressure 
existing  during  the  impact  can  be  satisfactorily  ascertained. 

Let  h  be  the  height  of  fall  above  the  top  of  the  rail,  and  d 
the  deflection  produced  by  P  falling  through  this  height.  Let 
R  be  the  mean  or  average  pressure  existing  between  the  rail 
and  weight  during  the  impact.  The  work  of  the  falling  ram  is 
P(h  -f-  £)  and  the  work  of  the  mean  pressure  R  is  R3.  Equat- 
ing these  values,  there  is  found 


from  which  R  can  be  computed  when  d  has  been  measured. 

This  formula  is  correct  both  for  elastic  and  non-elastic  de- 
flections. If  the  elastic  limit  be  not  exceeded,  d  can  be  com- 
puted by  (22)"  of  the  last  article.  If  the  elastic  limit  is 
exceeded,  8  can  only  be  found  by  actual  measurement. 

To  illustrate,  let  a  ram  weighing  2  ooo  pounds  fall  from  a 
height  of  20  feet  upon  a  railroad  rail,  this  being  one  method  of 
testing  rails  at  the  mill.  The  rail  will  be  deflected  an  amount 
#,  depending  upon  its  size,  shape,  length,  and  the  quality  of 
the  material.  For  8  =  i  inch  R  will  be  482  ooo  pounds  ;  for 
d  =  2  inches  R  will  be  242  ooo  pounds  ;  for  tf  =  4  inches  R 
will  be  1  22  ooo  pounds.  Thus  the  mean  pressure  decreases 
approximately  inversely  as  the  deflection. 

The  maximum  pressure  in  such  cases  cannot  be  theoreti- 
cally ascertained,  but  it  probably  is  not  as  great  as  double  the 
mean  pressure.  The  maximum  pressure  per  square  inch  will 
depend,  of  course,  upon  the  area  of  contact  between  the  ram 
and  the  rail,  and  upon  the  manner  in  which  the  total  pressure 
is  distributed  over  that  area. 

Prob.  157.  Compute  the  mean  pressure  on  the  rail  for  the 
data  given  at  the  foot  of  page  252. 


ART.  113.  CENTRIFUGAL  STRESS.  255 

ART.  113.    CENTRIFUGAL  STRESS. 

The  rod  that  connects  the  cross-head  of  an  engine  with  the 
crank  pin  is  subject  to  a  centrifugal  stress  owing  to  the  fact 
that  one  end  revolves  in  a  circle.  The  horizontal  rod,  or 
parallel  bar  joining  two  driving  wheels  of  a  locomotive  is  an- 
other instance  of  centrifugal  stress ;  this  is  simpler  than  the 
connecting  rod,  because  all  points  are  revolving  with  the  same 
velocity,  and  hence  it  will  be  discussed  first. 

Let  V  be  the  velocity  of  a  locomotive  in  feet  per  second, 
and  v  the  velocity  of  revolution  of  the  end  of  the  parallel  rod 
around  the  axle  of  the  driver  to  which  it  is  attached.  Let  R 
be  the  radius  of  the  driver,  and  r  the  radius  of  the  circle  of 
revolution  of  the  end  of  the  parallel  rod.  Then  since  the 


B  B 

FIG.  64. 

velocity  of  revolution  of  the  circumference  of  the  driver  is  the 
same  as  that  of  the  speed  of  the  train,  the  value  of  v  is 


Now,  not  only  the  end  but  every  point  in  the  parallel  rod  is 
revolving  with  the  velocity  v  in  a  circle  whose  radius  is  r. 
Thus  a  centrifugal  force  is  generated  which  produces  stresses. 
When  the  rod  is  at  its  lowest  position  BB,  this  centrifugal 
force  acts  as  a  downward  uniform  load  producing  flexure  ;  at 
the  highest  position  AA  it  acts  as  an  upward  uniform  load 
producing  flexure  ;  at  the  position  CC,  on  the  same  level  as 
the  axles,  it  produces  a  compressive  stress  in  the  direction  of 
the  length  of  the  rod. 


256  FLEXURE   OF  BEAMS.  CH.   IX. 

Let  w  be  the  weight  of  the  parallel  rod  per  linear  unit  ;  then 
from  mechanics  the  centrifugal  force  of  this  weight  is 


which  may  be  called  the  centrifugal  load  per  unit  of  length, 
The  rod  being  a  beam  supported  at  its  ends,  having  a  length  /, 
a  breadth  by  and  a  depth  d,  the  maximum  unit-stress  due  to 
this  uniform  load  is,  from  (4), 


__ 
/     ~  4&/8  ' 

which  is  the  flexural  stress  due  to  centrifugal  force  when  the 
bar  is  at  its  highest  or  at  its  lowest  position. 

In  this  formula  £*  is  the  acceleration  of  gravity,  or  32.16  feet 
per  second  per  second.  In  applying  it  numerically,  however, 
all  quantities  should  be  expressed  in  terms  of  the  same  linear 
unit,  the  inch  being  preferable.  For  example,  let  a  locomo- 
tive be  running  at  60  miles  per  hour,  the  radius  of  the  drivers 
being  3  feet  and  that  of  the  parallel  rod  I  foot,  this  being  of 
steel,  4  inches  deep,  2  inches  thick,  and  8  feet  long.  Here  V 
=  88  feet  per  second  =12  X  88  inches  per  second,  g  =  32.16 
X  12  inches  per  second  per  second,  R  =  3  X  12  inches,  r  = 
I  X  12  inches,  /  =  8  X  12  inches,  b  =  2  inches,  d  =  4  inches, 
and  w  =  2.27  pounds  per  linear  inch.  The  centrifugal  load 
per  inch  then  is 

.      2.27  X  12  X  (88  X  12)' 
f=  32.16  X  12  X  (3X12)-  =  6l  P°unds> 
and  then  the  maximum  fiber  stress  is 

e      3  X  61  X  (8  X  I2)a 

5  =  -  f.  -  =  13  200  pounds  per  square  inch, 

4  X  2  x   10 

which   is  not  probably  sufficiently  low  when  it  is  considered 
that  the  parallel  rod  is  subject  to  vibrations  and  shocks. 

The  connecting  rod  moves  in  a  circle  of  radius  r  at  the  crank 
pin,  while  the  other  end  moves  only  in  a  straight  line.  Thus 


ART.  114.  LIVE-LOAD  VELOCITY.  257 

at  the  end  A  there  is  no  centrifugal  load,  while  at  B  the  cen- 

trifugal load    is  the  same   as   given  by  the  above  expression 

for/".     When  the  rod  is  in  the 

position  shown  in  Fig.  65,  it  is  a 

beam  acted  upon  by  a  centrif- 

ugal   load    which    varies    uni- 

formly from  o  at  A  to  /  at  B. 

The  total  load  is  hence  \fl,  the  '        '     FlQ  6 

reaction  at  A  is  \fl  and  that  at 

B  is  4/7.     The  bending  moment  for  any  section  distant  x  from 

A  is 


and  the  maximum  value  of  M  occurs  for  x  =  //  1/3,  which 
gives  max.  M  =  0.0638/7*.     Hence  from  (4), 

//« 

^  =  0.383^,, 

in  which  f  is  given  by  the  same  expression  as  above. 

By  comparing  this  with  the  value  of  6"  for  the  parallel  rod  it 
is  seen  that  the  former  is  about  twice  as  great,  if  the  length 
and  cross-section  be  the  same  in  the  two  cases.  The  parallel 
rod  needs  the  greatest  cross-section  at  the  middle,  while  the 
connecting  rod  needs  the  greatest  cross-section  at  about  o.6/ 
from  the  cross-head. 

Prob.  158.  The  connecting  rod  of  an  engine  is  2  feet  long 
and  it  is  attached  to  a  crank  pin  at  a  distance  of  6  inches  from 
the  axis  of  a  fly-wheel.  If  the  wheel  makes  750  revolutions 
per  minute,  find  a  square  cross-section  for  the  connecting  rod 
so  that  the  centrifugal  unit-stress  5  may  be  4200  pounds  per 
square  inch. 

ART.  114.    LIVE-LOAD  VELOCITY. 

It  is  well  known  that  when  a  live  load  moves  over  a  beam 
or  bridge  the  deflections  and  stresses  are  greater  than  those 


258  FLEXURE  OF  BEAMS.  CH.  XI. 

due  to  the  same  load  at  rest.  In  general  the  greater  the  veloc- 
ity the  greater  also  are  the  deflections  and  the  stresses.  The 
exact  theoretical  investigation  of  this  question  is  one  of  very 
great  difficulty,  as  differential  equations  arise  which  cannot  be 
integrated  except  by  an  unsatisfactory  tentative  process. 
Approximate  formulas  have,  however,  been  established,  and  one 
of  these  will  here  be  deduced. 

Let  a  static  load  P  rest  on  the  middle  of  a  simple  beam. 
From  Chapter  III  the  deflection  and  the  maximum  unit-stress 
under  the  load  are 


A  C  - 

~'  :         ' 


where  /  is  the  length  of  the  beam,  7  the  moment  of  inertia  of 
the  cross-section  with  respect  to  the  neutral  axis,  c  the  dis- 
tance from  that  axis  to  the  remotest  fiber  where  the  unit-stress 
is  S.  Now,  let  the  load  P  be  moving  horizontally  across  the 
beam  with  the  velocity  v,  and  when  it  reaches  the  center  let 
the  deflection  be  d  and  the  maximum  unit-stress  be  71  It  is 
required  to  find  $  and  T  in  terms  of  A  and  5. 

When  the  load  P  runs  over  the  beam,  the  curve  in  which  it 
moves  is  found  by  putting  kl=.  x  in  the  first  equation  on  page 
77,  or 


Differentiating  this  twice,  and  then  making  x  =  £/,  gives 
d*\  i          PI 


as  the  reciprocal  of  the  radius  of  curvature  of  this  curve  at 
the  middle  of  the  beam. 

It  is  assumed  that  when  P  reaches  the  middle  of  the  beam 
it  produces  the  same  deflection  A  as  if  it  were  at  rest,  and  also 
that  it  causes  a  downward  pressure  F  due  to  the  centrifugal 
force  arising  from  motion  in  the  curve.  This  pressure  F  gives 


ART.  114.  LIVE-LOAD  VELOCITY.  259 

rise  to  an  additional  deflection,  thus  increasing  A  to  d,  and  5 
to  r-    Then, 


Now,  to  find  TS  the  expression  for  centrifugal  force  is  known 
from  mechanics,  and  inserting  in  it  the  above  value  of  R, 
there  results 

/V  _    P'tv*       2P*lh 

~~          == 


in  which  i?/2g  has  been  replaced  by  h,  the  height  of  fall  which 
will  produce  v.  In  this  last  expression  P/EI  may  be  replaced 
by  its  value  in  terms  of  A  or  by  its  value  in  terms  of  5  from 
the  first  equations  given.  Making  these  substitutions  in  the 
formulas  for  tf  and  S,  they  become 


(23) 


which  give  the  approximate  deflection  and  maximum  unit- 
stress  at  the  middle  of  the  beam  due  to  a  load  P  moving  with 
the  velocity  v  =  ^/2gh. 

As  an  example  let  a  wrought-iron  plate  girder  have  a  span 
of  80  feet,  a  depth  of  7  feet  2  inches,  a  flange  cross-section  of 
38  square  inches,  and  a  moment  of  inertia  of  about  134000 
inches.  Let  it  be  required  to  find  the  deflection  and  maximum 
unit-stress  when  a  single  load  of  60  ooo  pounds  crosses  the 
girder  at  a  velocity  of  80  miles  per  hour.  Here  P  =  60  OOO 
pounds,  /=  960  inches,  7=  134000  inches,  c  =  43  inches, 
and  E  —  25  ooo  ooo  pounds  per  square  inch.  Then  when  the 
load  is  at  rest  at  the  middle  of  the  beam, 

^  =  °-33  inches,     S  =  4620  pounds  per  square  inch. 


26o  FLEXURE   OF  BEAMS.  CH.  XL 

Now,  a  speed  of  80  miles  per  hour  corresponds  to  a  velocity  of 
117  feet  per  second,  and  the  velocity  head  is 


Then  from  (23)  are  found  the  increased  deflection  and  unit- 
stress, 

6  =  0.33(1  +  0.029)  =  0.34  inches, 

T  =  4620(1  +  0.029)  =  4750  pounds  per  square  inch, 
which  show  the  influence  of  the  velocity  to  be  small. 

When  a  uniform  live  load  is  moving  over  the  beam  or  bridge, 
a  similar  investigation  may  be  made,  regarding  the  centrifugal 
force  at  each  point  as  a  vertical  load.  Let  w  be  the  uniform 
live  load  per  linear  unit  ;  then  when  this  extends  over  the 
whole  beam, 

5^/4 
~ 


87  ' 

are  the  static  deflection  and  maximum  fiber  stress  at  the  mid- 
dle. Let  8  be  the  deflection  and  T  the  unit-stress  when  the 
entire  uniform  load  is  moving  with  the  velocity  v,  and  let  h  be 
the  head  due  to  this  velocity.  Then  by  a  method  similar  in 
principle  to  the  above,  it  may  be  shown  that 


,1  + 

(23/ 


-X. 


which  are  the  approximate  deflection  and  unit-stress  at  the 
middle  of  the  beam  due  to  the  moving  load  wl. 

As  an  example  take  the  plate  girder  whose  data  are  given 
above  and  let  a  uniform  load  of  1800  pounds  per  linear  foot  be 
moving  over  it.  Then 

A  =  0.495  inches,  5=5  545  pounds  per  square  inch, 
are  the  static  deflection   and  unit-stress  at  the  middle.     As 
before,  h  =  213  feet,  and  then  from  (23)' 


ART.  114.  LIVE-LOAD   VELOCITY.  26l 

6  =  0.506  inches,  T  =  5  670  pounds  per  square  inch, 
which  are  only  about  2.2  per  cent  greater  than  the  static  values. 
It  should  be  remarked  that  the  allowance  made  for  velocity 
of  a  live  load  in  practice  is  much  greater  than  the  above  for- 
mulas indicate.  It  is  often  customary  to  add  from  10  to  30 
per  cent  to  the  statical  stresses  in  order  to  cover  the  effect  of 
velocity,  the  greater  values  being  used  for  the  shorter  bridges. 
It  should  also  be  said  that  experiments  indicate  a  greater 
increase  in  deflection  than  these  formulas  give.  The  most  ex- 
tensive set  of  experiments  in  this  direction  is  that  made  by 
JAMES,  WILLIS,  and  GALTON,  for  the  British  board  of  1848, 
and  in  some  of  these  the  statical  deflection  was  more  than 
doubled  under  heavy  loads. 

It  may  be  further  noted  that  a  perfect  formula  for  the  effect 
of  velocity  of  live  load  would  show  in  the  case  of  very  high 
speeds  that  there  would  be  no  increase  in  deflection,  since 
there  would  then  not  be  sufficient  time  for  the  load  to  fall 
through  the  distance  tf.  This  was  recognized  by  the  board 
above  mentioned,  and  the  fact  ascertained  in  several  tests. 
For  instance  a  wrought-iron  beam  9  feet  long,  I  inch  wide,  and 
3  inches  deep  was  subjected  to  a  load  of  1778  pounds  moving 
at  different  velocities,  with  the  following  results : 

Velocity  in  feet  per  second,     o        15       29       36      43, 
Deflection  in  inches,  0.29    0.38    0.50   0.62    0.46. 

Here  it  will  be  seen  that  the  deflection  for  43  feet  per  second 
is  less  than  that  for  36  feet  per  second.  Unfortunately  these 
experiments  were  made  without  regard  to  the  elastic  limit  of 
the  material,  and  hence  the  results  are  of  little  use  as  a  test 
of  theory;  for  instance,  the  static  load  of  1778  pounds  causes 
a  unit-stress  of  32  ooo  pounds  per  square  inch  on  the  middle  of 
the  wrought-iron  beam,  and  thus  at  all  velocities  the  elastic 
limit  was  surpassed. 

The  formulas    of   this  article  are  confessedly  imperfect,  as 


262  FLEXURE   OF  BEAMS.  CH.  XL 

they  do  not  take  into  account  the  influence  of  the  inertia  of 
the  beam  which  will  tend  to  modify  them  materially.  This 
very  complex  question  cannot  here  be  investigated,  but  the 
student  is  referred  to  Appendix  B  of  the  Report  of  the  Com- 
missioners on  the  Application  of  Iron  to  Railway  Purposes 
(London,  1849)  f°r  an  excellent  general  discussion.  It  may 
also  be  noted  that  the  above  formula  (23)  agrees  with  the  first 
two  terms  of  the  series  given  on  page  203  of  that  Report. 

Prob.  1 59.  Deduce  the  values  of  $  and  T  given  in  formula 
(23)'  for  the  uniform  moving  load. 

Prob.  160.  What  velocity  v  must  the  load  P  have  so  that, 
in  crossing  the  above  plate-girder,  there  would  not  be  sufficient 
time  for  it  to  fall  through  the  vertical  deflection  of  0.33  inches  ? 

ART.  115.    WORK  OF  VERTICAL  SHEARS. 

In  the  discussion  of  Art.  109  the  entire  external  work  of 
the  load  P  was  supposed  to  be  expended  in  the  work  of  elon- 
gating and  compressing  the  horizontal  fibers  of  the  beam.  In 
reality,  however,  a  part  of  the  external  work  is  expended  in 
the  slipping  or  detrusion  due  to  the  vertical  shears  throughout 
the  beam. 

As  in  Chapter  III  suppose  the  vertical  shear  to  be  uniformly 
distributed  over  the  cross-section  of  the  beam. 
Let  Fbe  the  vertical  shear  at  any  distance  x  from 
an  origin  and  0  be  the  angle  of  detrusion  in  the 
distance  dx.  Then  taking  the  shear  to  increase 
slowly  from  o  up  to  its  value  V,  the  work  done  by 
it  in  the  distance  dx  is,  since  0  is  very  small, 

dK=\V.dx.  tan  0  =  %V(J)dx, 
in  which   0  in  the  last  value  is  to  be  taken  in 
circular  measure. 

FIG.  66.  Let  A  be  the  cross-section  of  the  beam,  5S  the 

shearing  unit-stress,  and   Es  the   coefficient   of   elasticity  for 


•dx 


ART.  US.  WORK  OF  VERTICAL   SHEARS.  263 

shearing.  As  0  is  the  amount  of  detrusion  per  unit  of  dis- 
placement, its  value  is  found  by  the  same  process  as  the  unit- 
deformation  s  for  tension  or  compression,  namely 

s,      v 
*••--£.=--  ~AE; 

This  being  inserted,  the  expression  for  dK  becomes 

V'd* 


which  is  the  elementary  work  of  shearing  in  the  distance  dx. 
By  expressing  V  as  a  function  of  x  and  integrating  over  the 
entire  length  of  the  beam,  the  total  work  of  shearing  is  found. 

For  instance,  a  simple  beam   loaded  with  P  at  the  middle 
has  the  shear  V  constant  throughout  and  equal  to  \P.    Then 

K  > 

= 


is  the  internal  work  or  resilience  due  to  all  the  shearing  forces 
over  the  span  /.  In  Art.  109  the  internal  work  of  the  horizon- 
tal stress  was  found  to  be 


so  that  the  ratio  of  the  former  to  the  latter  is 
K^        12EI  Ef 

K' 


in  which  r  is  the  radius  of  gyration  of  the  cross-section  with 
respect  to  the  neutral  axis. 

For  example,  let  a  cast-iron  beam  have  a  square  cross-section 
of  side  d\  then  ra  =  TVA  and  E  =  %ES  approximately  (Art. 
121).  The  ratio  of  the  internal  work  of  shearing  to  that  of 
the  fiber  stresses  is  then  5*/a/2/3.  If  the  length  is  30  times 
the  depth,  or  /  =  30^,  then  this  ratio  is  -^  ;  if  /  =  6o*/,  the 
ratio  is  ±  4*4  0  .  For  a  very  short  beam,  such  as  /  =  2d,  the 
ratio  is  £  ,  showing  that  in  such  cases  the  work  of  shearing  is 
greater  than  that  of  the  horizontal  fiber  stresses. 


2b4  FLEXURE   OF   BEAMS.  CH.  XI. 

Prob.  161.  Deduce  an  expression  for  the  work  of  shearing  in 
a  beam  supported  at  the  ends  and  uniformly  loaded. 

ART.  116.    DEFLECTION  DUE  TO  SHEARING. 

The  treatment  of  the  deflection  of  beams  in  the  previous 
pages  has  been  solely  from  the  standpoint  of  the  horizontal 
stresses  as  expressed  by  the  external  bending  moment.  The 
last  Article  shows,  however,  that  the  resisting  work  of  shearing 
may  be  a  material  amount  for  short  beams,  and  it  hence 
appears  that  the  former  investigations  are  more  or  less  incom- 
plete. 

Let  a  load  P  produce  the  deflection  A  beneath  it.  The  ex- 
ternal work,  if  P  has  been  gradually  applied,  is  %P4,  and  this 
must  equal  the  internal  work,  or  resilience,  of  the  molecular 
stresses.  The  work  of  the  horizontal  stresses  of  tension  and 
compression  is  deduced  in  Art.  109  and  that  of  the  vertical 
stresses  of  shearing  in  Art.  115.  Then  the  [external  work 
equated  to  the  sum  of  these,  gives 

C M*dx   .       CVdx 
J 


in  which  M  is  the  bending  moment  and  V  the  vertical  shear  at 
any  section  distant  x  from  the  origin.  To  apply  this  to  a  par- 
ticular case,  M  and  V  are  to  be  expressed  in  terms  of  x  and 
the  integration  extended  over  the  entire  length  of  the  beam. 

For  example,  let  a  simple  beam  of  span  /  have  a  load  P  at 
the  middle.  Then  M=  \Px  and  V  =  \P.  Inserting  these 
and  bearing  in  mind  that  each  integral  is  equal  to  twice  the 
value  between  the  limits  o  and  £/,  there  is  found, 

A-      Pr      .      Pl 
~48£7  ~h4^' 

which  is  the  deflection  under  the  load.  The  second  term  here 
gives  the  deflection  due  to  the  vertical  shears.  By  placing 


ART.  1  1  6.  DEFLECTION   DUE  TO   SHEARING.  265 

A  =  7/ra,  where  r  is  the  radius  of  gyration,  this  becomes 

= 


which  is  the  formula  for  deflection,  taking  into  account  the 
effect  of  the  vertical  shears.  For  long  beams  the  deflection 
due  to  shearing  is  scarcely  appreciable  ;  for  short  beams,  how- 
ever, it  may  be  larger  than  that  due  to  the  bending  moment. 

Attention  was  first  called  to  the  inaccuracy  of  the  ordinary 
formula  for  deflection  in  the  case  of  short  beams  in  a  paper  by 
NORTON  read  before  the  American  Association  for  the  Ad- 
vancement of  Science  in  1870.  A  number  of  experiments  on 
white  pine  beams  of  different  lengths  and  sizes  were  made,  and 
it  was  shown  that  the  deflections  were  directly  proportional  to 
the  loads  and  inversely  proportional  to  the  breadth  of  the 
beam,  as  the  common  formula  requires.  The  deflections  were, 
however,  not  directly  proportional  to  the  cubes  of  the  spans 
nor  inversely  proportional  to  the  cubes  of  the  depths  of  the 
beams,  as  the  formula  requires.  An  examination  into  the 
reason  of  these  discrepancies  showed  that  it  was  due  to  influence 
of  the  vertical  shears,  and  NORTON  deduced  the  formula 

~~ 

as  applicable  to  beams  of  breadth  b  and  depth  d,  where  C  was 
a  constant  whose  value  he  did  not  theoretically  determine. 
From  one  series  of  experiments  he  found  for  the  white  pine 
beams 

E  =  1  428  ooo,    C  =  0.0000094, 

and  using  these  values  the  deflections  for  other  series  com- 
puted from  the  formula  agreed  very  well  with  those  observed. 

From  the  theoretic  formula  for  A  deduced  above,  it  is  seen 
that  NORTON'S  value  of  C  is  1/4  Es,  or 

Es  =  —  ^  =  266  ooo  pounds  per  square  inch, 
4C 


266  FLEXURE  OF  BEAMS.  CH.  XL 

which  should  be  the  coefficient  of  elasticity  for  the  shearing  of 
white  pine  across  the  grain.  This  is  probably  not  far  from  the 
actual  value  of  that  coefficient,  since  THURSTON,  by  experi- 
ments on  torsion,  found  Es  =  220  ooo  pounds  per  square  inch 
for  white  pine.  The  experiments  of  NORTON,  therefore,  con- 
firm the  theoretic  formula  above  deduced  for  the  true  deflec- 
tion of  a  simple  beam  loaded  at  the  middle. 

When  a  uniform  load  extends  over  the  beam  and  it  is  de- 
sired to  find  the  deflection  at  a  particular  point,  let  M  and  V 
be  the  bending  moment  and  vertical  shear  due  to  the  uniform 
load,  and  M'  and  V  the  bending  moment  and  vertical  shear 
due  to  a  load  P'  placed  at  the  particular  point.  Then  the 
deflection  at  that  point  is  given  by 

CV'Vdx 


I     ^JP'E^ 

in  which  both  bending  moments  and  vertical  shears  are  to  be 
expressed  in  terms  of  x,  and  the  integration  extended  over 
the  entire  length  of  the  beam.  For  instance,  let  a  simple 
beam  have  the  uniform  load  wl,  and  let  it  be  required  to  find 
the  deflection  at  the  middle.  Here  M'  =  fP  'x,  M=\w(lx  — 
x*\  V'  =  \P' >  and  V '=  \wl  —  wx.  Inserting  and  integrating 
between  the  limits  o  and  /,  there  results 

wl* 


which  is  the  deflection  at  the  middle,  the  first  term  being  the 
deflection  due  to  the  horizontal  stresses  of  tension  and  com- 
pression and  the  second  that  due  to  the  vertical  stresses  of 
shearing.  The  ratio  of  the  second  term  to  the  first  is  seen  to 
be  slightly  greater  than  in  the  case  of  a  single  load  at  the 
middle. 

Prob.  162.  Prove  formula  (24)'  by  considering  that  the  work 
due  to  the  imaginary  load  P '  is  equal  to  the  stress  caused  by 


ART.  II/.  FLEXURE  AND   COMPRESSION.  267 

that  load  multiplied  by  the  deformation  caused  by  the  stress 
due  to  the  total  uniform  load. 

ART.  117.     FLEXURE  AND  COMPRESSION. 

Let  a  beam  be  subject  to  flexure  by  transverse  loads  and 
also  to  a  compression  in  the  direction  of  its  length.  If  the 
longitudinal  compression  be  not  large  the  combined  maximum 
stress  due  to  flexure  and  compression  may  be  computed  by 
the  approximate  method  of  Art.  74.  It  is  clear,  however, 
that  if  the  compression  be  large  the  deflection  A  will  be 
increased,  and  hence  the  effective  bending  moment  and  maxi- 
mum fiber  stresses  will  be  greater  than  given  by  that  method. 
A  closer  approximation  will  now  be  established. 

Let  P  be  the  longitudinal  compressive  force  and  M  the 
bending  moment  of  the  flexural  forces.  Let  Ml  be  the  actual 
bending  moment  for  the  section  where  the  deflection  is  A  ; 
this  is  greater  than  M,  on  account  of  the  moment  PA  of  the 
force  P,  or  M^  =  M  -{-  PA.  Now  the  maximum  fiber  unit- 
stress  5,  which  results  from  this  moment  Ml  is,  from  (4), 
_  M,c  _  (M+PA)c 

•*l  =     /  7~  "' 

where  I  is  the  moment  of  inertia  of  the  cross-section  and  c 
the  distance  from  the  neutral  axis  to  the  remotest  fiber. 
The  value  of  A  may  be  expressed  in  terms  of  5",  regarding  A 
to  vary  with  Sj  in  the  same  manner  as  for  a  beam  subject  to 
no  longitudinal  compression.  Inserting  then  for  A  its  value 
from  Art.  37,  and  solving  for  S19  gives 

Me 
~= 


_ 

mE 

where  n  and  m  are  numbers  depending  upon  the  arrangement 
of  the  ends  and  the  kind  of  loading.  This  formula  was  first 
deduced  by  J.  B.  JOHNSON,  who  regarded  m/n  as  10  for  all 
kinds  of  loading.  Art.  37  shows,  however,  that  m/n  depends 


268  FLEXURE  OF  BEAMS.  CH.  XL 

on  the  arrangement  of  the  ends  as  well  as  on  the  load  ;  for  a 
simple  beam  uniformly  loaded  m/n  =  9.6,  and  for  a  load  at 
the  middle  m/n  =12. 

The  maximum  compressive  unit-stress  on  the  concave  side 
of  the  beam  is  5  =  5X  -|-  P/A.  For  example,  let  a  wooden 
beam  8  feet  long,  10  inches  wide,  and  9  inches  deep  be  under 
a  compression  of  40  ooo  pounds,  while  at  the  same  time  it 
carries  a  total  uniform  load  of  4000  pounds.  Here  M  =  \WL 
—  48  ooo  pound-inches,  c  =  4^  inches,  /  =  -r^bd* 
inches4,  /  =  96  inches,  P  =  40  ooo  pounds,  n  =  8,  m  = 
and  E  =  I  500  ooo  pounds  per  square  inch.  Inserting  these 
values  in  the  formula  the  flexural  stress  Sl  is  found  to  be  371 
pounds  per  square  inch.  The  compressive  unit-stress  due 
directly  to  P  is  P/A  =  40  000/90  =  444,  so  that  the  total 
stress  5  =  371  -{-  444  =815  pounds  per  square  inch. 

While  the  above  method  is  better  than  that  of  Art.  74,  it 
is  not  exact,  and  gives  in  general  values  of  5  which  are  too 
small.  The  exact  method  of  dealing  with  combined  flexure 
and  compression  is  by  the  help  of  the  elastic  curve.  The 
results  will  now  be  developed  for  the  most  common  case. 

Let  a  simple  beam  of  span  /  be  uniformly  loaded  with  in 
per  linear  unit,  and  at  the  same  time  be  under  the  longitudinal 
compression  P.  The  bending  moment  for  any  point  whose 
co-ordinates  are  x  and  y  is, 

M  =  \wlx  —  %wx*  +  />, 
and  the  differential  equation  of  the  elastic  curve  is, 


where  the  negative  sign  of  the  bending  moment  is  taken 
because  the  curve  is  concave  to  the  axis  of  x.  By  two 
integrations  results 

wlx       wx*       wEI  /cos  fi(x  —  J/)         \ 
~~2P  "  ~2P  +  ~7*~\  ~         Vf 


ART.  117.  FLEXURE  AND  COMPRESSION.  269 

in  which  /?,  as  in  Art.  62,  is  an  abbreviation   for  (P/EI}*,  or 


is  an  arc  expressed  in  terms  of  the  radius  as  unity.     In  this 
equation  of  the  elastic  curve  let  x  =  i/,  then^  =  J,  and 


Inserting  this  in  the  expression  for  St  ,  there  is  found 


which  is  an  exact  expression  for  the  maximum  compressive 
flexural  unit-stress.  Lastly,  5,  -f-  P/A  is  the  total  unit-stress 
5  due  to  the  combined  flexure  and  compression. 

To  illustrate  this  method  let  the  data  of  the  above  numeri- 
cal example  be  again  used.  Here  w  =  4000/96  pounds  per 
linear  inch,  and  the  other  quantities  as  before.  The  arc  £/?/ 
is  found  to  be  0.318,  and  the  corresponding  angle  is  18°  13', 
whence  sec  \ftt  =  1.0545.  Then  from  the  formula  the 
flexural  unit-stress  Sl  is  416  pounds  per  square  inch.  Lastly, 
the  total  unit-stress  is  416  +  444  =  860  pounds  per  square 
inch.  A  comparison  for  this  numerical  example  shows  that 
the  rough  method  of  Art.  74  gives  799,  and  the  method  of 
JOHNSON  815,  while  the  exact  method  gives  860  pounds  per 
square  inch  for  the  maximum  unit-stress. 

If  w  =  o  in  the  above  formulas  the  case  reduces  to  that  of 
a  column  where  sec  i/?/  =  oo  ,  and  hence  both  A  and  5,  are 
indeterminate.  If  on  the  other  hand  P  =  o,  the  case  is  that 
of  a  simple  beam  uniformly  loaded,  and  it  may  be  shown  that 
A  and  5,  will  reduce  to  the  expressions  for  that  case. 

Prob.  163.  Show,  by  the  calculus  method  of  evaluating 
indeterminate  quantities,  that  the  statement  in  the  last  sen- 
tence is  correct. 


270  FLEXURE   OF  BEAMS.  CH.  XL 

Prob.  164.  Let  a  simple  wooden  beam  32  feet  long,  9 
inches  wide,  and  10  inches  deep,  carry  a  total  uniform  load  of 
2000  pounds  while  at  the  same  time  it  is  under  a  longitudinal 
compression  of  9000  pounds.  Compute  the  maximum  unit- 
stress  »S  by  the  three  methods. 

ART.  118.     FLEXURE  AND- TENSION. 

Let  a  beam  be  subject  to  flexure  by  transverse  loads  and 
then  to  a  tension  in  the  direction  of  its  length.  The  effect 
of  the  tension  is  to  decrease  the  deflection,  and  thus  also  the 
tensile  flexural  stress.  If  M  be  the  bending  moment  of  the 
transverse  loads,  and  Ml  that  of  the  combined  flexure  and 
tension,  then  M^  =  M  —  PA.  Let  S,  be  the  resulting  unit- 
stress  on  the  fiber  most  remote  from  the  neutral  surface; 
then,  JOHNSON'S  formula  in  the  last  article  gives  S, ,  if  the 
minus  sign  in  the  denominator  be  changed  to  plus.  Finally, 
S,  -f-  PI -A  is  the  total  unit-stress  on  the  convex  side  of  the 
beam  resulting  from  the  combined  flexure  and  tension. 

As  an  example  take  a  steel  eye-bar  18  feet  long,  I  inch 
thick,  and  8  inches  deep,  under  a  longitudinal  tension  of 
80000  pounds,  E  being  29000000  pounds  per  square  inch. 
The  weight  of  the  bar  is  490  pounds,  and  M  =  j-  X  490  X  18 
X  12  =  13  230  pound-inches.  Also  c  =  4  inches,  /=  42.67 
inches4,  m/n=  9.6,  P  =.  80000  pounds,  /=  216  inches. 
Then  the  maximum  flexural  tensile  stress  Sl  is  943  pounds 
per  square  inch.  Finally,  the  total  tensile  stress  is  5  =  943 
-f-  10  ooo  =  10  943  pounds  per  square  inch. 

As  in  the  last  article,  a  more  accurate  way  of  dealing  with 
this  case  is  by  use  of  the  general  equation  of  the  elastic  curve. 
The  expression  for  the  bending  moment  is, 
M  =  \wlx  —  \ivx*  —  />, 

which  is  the  same  as  before,  except  in  the  sign  of  P.  Hence 
by  changing  the  sign  of  P  in  the  expressions  for  A  and  5, , 
they  apply  to  the  case  of  combined  flexure  and  tension.  In 


ART.  Il8.  FLEXURE  AND  TENSION. 


doing  this  \fil  becomes  ^filV--  i,  and  this  changes  the  cir- 
cular secant  to  the  hyperbolic  secant;  thus, 


is  the  deflection  of  the  beam,  and 

S,  =  ~(i  -  sech 

is  the  unit-stress  due  to  the  flexure.     Finally  St  +  P/A  is  the 
total  unit-stress  due  to  the  combined  flexure  and  tension. 

Since  many  students  are  unacquainted  with  hyperbolic 
functions,  it  may  be  here  noted  that  they  are  closely  analogous 
with  the  circular  functions.  Thus  for  circular  functions  cos* 
-f-  sina  =  i,  but  for  hyperbolic  functions  cosh'  —  sinh"  =  I. 
A  table  of  hyperbolic  sines,  cosines,  and  tangents  is  given  in 
"  Higher  Mathematics"  (Wiley  &  Sons,  1896).  In  the 
absence  of  a  table  the  hyperbolic  cosine  and  secant  can  be 
computed  from 

e6  +  e~  e 
cosh  6  r       —  -  ,     sech  0  = 


where  e  is  the  base  of  the  Naperian  system  of  logarithms. 

As  an  example,  for  the  above  eye-bar  w  =  2.267  pounds 
per  linear  inch,  and  i/?/  =  0.868  =  #;  then  sech  6  =  0.714, 
and  the  flexural  stress  S,  =  940  pounds  per  square  inch. 
Lastly,  the  total  unit-stress  5  =  940  -f-  10000  =  10940 
pounds  per  square  inch,  which  differs  but  little  from  the  result 
found  before. 

Prob.  165.  Compute  the  deflection  A  for  the  above  eye- 
bar  before  and  after  the  tension  is  applied. 

Prob.  166.  Show  that  the  deflection  of  a  cantilever  beam 
loaded  at  the  end  with  W  and  under  the  longitudinal  tension  P 
is  J  =  WP-\l  -  ft'1  tanh  /?/),  where  ft  =  (//£/)*. 


272 


SHEAR   AND   TORSION. 


CH.  XII 


CHAPTER   XII. 
SHEAR  AND  TORSION. 

ART.  119.    STRESSES  CAUSED  BY  SHEAR. 

It  is  shown  in  Art.  8,  and  also  in  Art.  75,  that  forces  of  ten- 
sion or  compression  acting  upon  a  body  produce  not  only 
internal  tensile  or  compressive  stresses,  but  also  internal  shear- 
ing stresses.  Conversely,  an  external  shear  acting  upon  a 
body  produces  in  it  not  only  internal  shearing  stresses,  but  also 
internal  tensile  and  compressive  stresses. 

For  example,  the  rectangle  ABCD  in  the  web  of  a  plate 
girder,  shown  in  Fig.  67,  may  be  considered.  Let  V  be  the 
shear  at  the  sections  AB  and  CD,  which 
are  taken  very  near  together  so  that  the 
weight  in  the  rectangle  itself  can  be  dis- 
regarded. This  vertical  shear  or  couple 
must  be  accompanied  by  a  horizontal 
shear  F,,  which  in  this  case  is  caused  by 
the  resistance  of  the  flange  >  rivets.  Let 
the  thickness  of  the  material  be  one  unit ; 
then  if  5  and  Sl  be  the  shearing  unit- 
stresses, 

—    A  f}y  i   —    A  f~V 

^1Z>  ./I// 

and  it  is  now  to  be  shown  that  5  and  5X  are  equal.  Taking 
either  A  or  D  as  a  center  of  moments,  the  equation  of  mo- 
ments is, 

V  X  AD  =  V,  X  AB, 


FIG.  67. 


ART.   119.  STRESSES   CAUSED   BY   SHEAR.  2/3 

and  hence  by  division 

V         V 

AB  =  AD>     °r     5=5» 

that  is,  the  shearing  unit-stresses  on  adjacent  sides  of  the  rec- 
tangle are  equal.  This  is  without  regard  to  the  weight  of  the 
rectangle  itself,  which  will  cause  a  slight  modification,  because 
the  V  on  the  left  will  then  be  greater  than  the  V  on  the  right. 
But  if  AD  be  very  small  the  conclusion  is  strictly  true  that  the 
vertical  shearing  unit-stress  is  equal  to  the  horizontal  shearing 
unit-stress  (Art.  79). 

The  resultant  of  V  and  F,  acts  as  a  tension  on  the  diagonal 
BD  and  as  a  compression  on  the  diagonal  AC,  thus  tending  to 
deform  the  rectangle  into  a  rhombus.  The  maximum  value  of 
this  resultant  will  be  when  F,  =  F,  that  is,  when  the  rectangle 
is  a  square..  The  resultant  tension  or  compression  then  acts 
at  an  angle  of  45  degrees  with  the  length  of  the  beam,  and  its 
value  is  VjH.  The  tensile  or  compressive  unit-stress  is  ob- 
tained by  dividing  F\/2  by  the  area  normal  to  its  direction, 
or 


that  is,  the  tensile  or  compressive  unit-stress  caused  by  shear 
is  equal  to  the  shearing  unit-stress. 

This  may  also  be  proved  from  the  discussion  in  Art.  75. 
Thus  in  formula  (13)  let  /  =  o;  then  max.  /  =  ±  vy  which  is 
the  same  result.  The  action  of  tension  or  compression  on  a 
bar  produces  a  shearing  unit-stress  equal  to  one  half  the  tensile 
or  compressive  unit-stress,  but  the  action  of  a  shear  \  oduces 
tensile  and  compressive  unit-stresses  equal  to  the  sheai  ng  unit- 
stress  itself.  This  may  be  regarded  as  a  most  fortunate  arrange- 
ment in  view  of  the  fact  that  the  shearing  strength  •  f  materi- 
als is  usually  less  than  the  tensile  or  compressive  strength. 

Prob.    167.  A    square  bar  is  subject  to  a  tension    of   6000 


274  SHEAR   AND   TORSION.  CH.  XII. 

pounds  per  square  inch  in  the  direction  of  its  length  and  to  a 
lateral  compression  of  2000  pounds  per  square  inch  on  two 
opposite  sides.  Show  that  the  maximum  shearing  unit-stress 
in  the  bar  is  4000  pounds  per  square  inch. 

ART.  120.    RESILIENCE  UNDER  SHEAR. 

The  resilience  of  a  body  under  the  action  of  shear  is  gov- 
erned by  similar  laws  to  that  of  tension  and  flexure,  namely,  it 
is  proportional  to  the  square  of  the  maximum  unit-stress  and  to 
the  volume  of  the  body.  Thus  in  Fig.  68  let  a  shearing  unit- 
stress  Sg  act  upon  a  parallelepiped  of  length  /  and  cross-section 
A,  deforming  it  into  a  rhombus  and  causing  each  right  angle 
on  the  side  to  be  increased  or  decreased  by 
the  amount  0.  The  external  work  done 
by  the  total  shear  V,  supposing  it  be 
gradually  applied,  is  £  VI  tan  0,  or  since  0 
is  a  very  small  angle,  simply  •£  F/0,  if  0 
be  expressed  in  circular  measure.  This  is 
equal  to  the  internal  work  or  resilience  of 
FIG.  68.  all  the  shearing  stresses.  But  F=  AS,, 

and  if  Et  be  the  coefficient  of  elasticity  for  shearing,  then 
5,  =  £,0(Art.  4).  Therefore 


is  an  expression  for  the  work  of  shearing.  The  first  factor 
.St/2Et  may  be  called  the  modulus  of  resilience  for  shearing, 
in  analogy  with  that  for  tension  (Art.  94),  and  the  second  fac- 
tor Al  is  the  volume  of  the  body.  This  expression,  however, 
is  only  valid  when  the  stress  Ss  is  within  the  elastic  limit  of  the 
material. 

Practically  a  shear  cannot  act  upon  a  body  of  any  consider- 
able size  without  causing  flexure  or  torsion,  and  thus  only  a 
part  of  the  external  work  will  be  expended  in  the  internal 
work  of  shearing.  Hence  the  above  rule  for  resilience  under 


ART.   121.         THE   COEFFICIENTS   OF  ELASTICITY. 


275 


shear  is  of  limited  application,  unless  the  effect  of  the  accom- 
panying flexure  be  considered.  In  the  case  of  long  beams  the 
work  of  shearing  is  indeed  but  a  small  part  of  that  due  to  the 
flexure  (Art.  115). 

The  ultimate  resilience  of  shearing  is  far  more  difficult  to 
estimate  than  that  of  tension  or  flexure.  It  can,  however,  be 
experimentally  determined  by  the  power  required  to  punch  a 
hole  through  a  plate,  although  even  in  this  case  some  of  the 
applied  work  is  lost  in  heat  and  friction. 

Prob.  168.  Estimate  the  horse-power  required  to  punch  50 
holes  per  minute  in  a  wrought-iron  plate  }  inch  thick,  the 
diameter  of  the  holes  being  2  inches. 

ART.  121.    THE  COEFFICIENTS  OF  ELASTICITY. 

The  coefficient  of  elasticity  for  shearing  has  a  certain  rela- 
tion to  the  coefficient  of  elasticity  for  tension,  which  will  now 
be  deduced.  Let  abed  represent  one  side  of  a  cube  which  un- 
der the  tensile  unit-stress  5  is  elongated  into  the  parallelepiped 

£  o 

D 


FIG.  69. 

A  BCD,  the  length  ab  being  increased  to  AB,  and  the  breadth 
ad  being  decreased  to  AD.  The  ratio  of  the  lateral  decrease 
to  the  longitudinal  increase  is  designated  by  e,  a  mean  value  of 
which  for  iron  and  steel  is  £,  as  already  mentioned  in  Art.  71. 
The  distance  AB  —  ab  is  the  unit-elongation  s,  and  the  distance 
ad  —  AD  is  the  lateral  unit-contraction  es. 

The  distortion  of  the  square  into  the  parallelogram  may  be 
regarded  as  caused  by  the  shearing  stresses  acting  along  the 


2/6  SHEAR  AND   TORSION.  CH.  XII. 

two  diagonals  AC  and  BD.  The  angle  cab,  originally  JTT,  is 
changed  into  CAB,  which  is  \n  —  J0  ;  the  total  change  of 
angle  between  the  two  diagonals  of  abed  being  the  distortion 
0  due  to  shearing.  Now 

BC        i  —  es 


, 

and  since  0  is  very  small,  the  value  of  tan  (J?r  —  £0)  is  i  —  0. 
very  nearly.  Hence 

I    —  GS 

i  —  0  =  Tqrj'       or>       0  =  (r  +  e)s> 

since  after  reduction  s  can  be  neglected  in  comparison  with  I. 
Lastly,  replacing  for  0  its  value  SS/ES  ,  and  for  s  its  value  S/E, 
and  remembering  that  Ss  =  %S,  as  shown  in  Art.  7,  there  is 
found  the  important  formula, 


'       2(+)' 

which  gives  the  coefficient  of  elasticity  for  shearing  in  terms  of 
the  coefficient  of  elasticity  for  tension. 

The  abstract  number  e  is  called  the  factor  of  lateral  contrac- 
tion. For  cast  iron  its  value  is  found  to  be  about  i,  and  thus 
Es  =  f  E.  For  wrought  iron  and  steel  e  is  about  £,  and  for 
these  materials  Es  =  %E.  For  fibrous  or  non-homogeneous 
materials,  however,  formula  (25)  often  fails  to  apply,  for  the 
reason  that  E  is  not  the  same  in  all  directions  as  it  is  in  a 
homogeneous  body.  Using  the  mean  values  of  E  given  in 
Art.  80,  the  mean  values  of  the  coefficients  of  elasticity  for 
shearing  are, 

For  cast  iron,  Es  =  6  ooo  ooo, 
For  wrought  iron,  Es  =  9  400  ooo, 
For  steel,  £,=  11  200  ooo, 

all  being  in  pounds  per  square  inch.  By  means  of  experi- 
ments on  the  torsion  of  shafts  (Art.  66)  these  values  have  been 
verified. 


ART.   122.  RESILIENCE    UNDER  TORSION.  277 

Prob.  169.  Prove  that  tan  (\TT  —  £0)  is  I  —  0  very  nearly, 
when  0  is  small. 

Prob.  170.  A  cast-iron  shaft  60  inches  long  and  2  inches  in 
diameter  is  twisted  through  an  angle  of  7  degrees  by  a  force  of 
2500  pounds  acting  at  12  inches  from  the  center,  and  on  the 
removal  of  the  force  springs  back  to  its  original  position. 
Compute  the  factor  of  lateral  contraction  e. 

ART.  122.    RESILIENCE  UNDER  TORSION. 

When  a  shaft  is  twisted  by  a  force  P  acting  with  a  lever  arm 
/»,  as  in  Fig.  49  of  Art.  63,  each  element  of  the  cross-section  is 
subject  to  a  shearing  unit-stress  S.  The  stress  being  slowly 
developed,  the  internal  work,  or  resilience,  is  equal  to  \S  mul- 
tiplied by  its  displacement  0,  or  if  Et  denote  the  coefficient  of 
elasticity  for  shearing,  the  work  of  any  elementary  area  a  and 
length  dx  is 

dK  =    5    .adx=l-  adx. 


Now  let  St  be  the  shearing  unit-stress  at  the  part  of  the  cross- 
section  most  remote  from  the  axis,  and  let  c  be  its  distance 
from  that  axis  ;  also  let  z  be  the  distance  of  S  from  the  axis. 

Then  5  =  S,  -,  and 

.„       i  S;  a* 

dK  =  -  -=-  .  —  r  .  dx 
2  Et     c* 

is  an  expression  for  the  internal  work.  To  integrate  this  over 
the  entire  volume  of  the  shaft  all  cross-sections  being  similar, 
2  at?  is  the  polar  moment  of  inertia  J,  and  2dx  is  the  length 
of  the  shaft  /.  Thus 

/^\  r       l  S*    J  i       I  S*   ^  AJ 

'"iy-p'-ii-?^* 

where  r  is  the  polar  radius  of  gyration  of  the  cross-section  de- 
fined byy  =  Ar*.  Now  Al  is  the  volume  of  the  shaft,  and  it 
is  thus  seen  that  the  resiliences  of  shafts  of  similar  cross-sec- 


SHEAR  AND  TORSION.  CH.  XIL 

tions  are  proportional  to  their  volumes.  Hence  the  resilience 
of  a  shaft  under  torsion  is  governed  by  laws  similar  to  those  of 
a  beam  under  flexure  (Art.  96). 

The  formula  here  established  is  only  valid  when  the  greatest 
unit-stress  55  does  not  surpass  the  elastic  limit  for  shearing. 

1  >Sa 
When  .Ss  corresponds  to  the  elastic  limit,  the  quantity  -  ~  may 

2  •&* 

be  called  the  modulus  of  resilience  for  torsion  or  shearing,  in 
analogy  to  the  modulus  of  resilience  for  tension  or  compres- 
sion (Art.  94). 

As  an  example,  let  it  be  required  to  find  the  work  necessary 
to  strain  a  steel  shaft  12  inches  in  diameter  and  30  feet  long 
up  to  its  elastic  limit,  supposed  to  be  30000  pounds  per  square 
inch.  Here  5S  =  30000  and  Es  =  n  200000  pounds  per 
square  inch  (Art.  121);  also,  c  =  6  inches,  A  —  113.1  square 
inches,  /  =  A7^'  —  \Ad*  =  2O3^  inches4,  /  —  360  inches. 
Inserting  all  values,  K  is  found  to  be  818000  inch-pounds  or 
68  200  foot-pounds.  Thus  to  produce  this  stress  in  the  shaft 
in  one  minute  more  than  2  horse-powers  are  required. 

Prob.  171.  Compare  the  resilience  of  a  square  shaft  and  a 
round  shaft,  the  cross-sections  and  lengths  being  equal. 

ART.  123.    HOLLOW  AND  SOLID  SHAFTS. 

It  was  mentioned  in  Art.  69  that  a  hollow  shaft  is  stronger 
than  a  solid  shaft  of  the  same  sectional  area.  A  general  com- 
parison will  now  be  made  with  respect  to  strength,  stiffness,, 
and  resilience.  Let  A  be  the  ar«a  of  the  cross-section  in  both 
cases,  let  D  be  the  outer  and  d  the  inner  diameter  of  the  hol- 
low shaft;  then  A  =  \n(D*  —  d*\  and  the  diameter  of  the 
solid  shaft  is  d,  =  -/£>'—  d\ 

The  strength  of  a  shaft  under  torsion  is  measured  by  the 
twisting  moment  it  can  carry  under  a  given  unit-stress,  and  by 


ART.  123.  HOLLOW  AND   SOLID  SHAFTS.  279 

Art.  64  this  is  seen  to  vary  as  J/c,  or  as  its  polar  moment  of 
inertia  divided  by  its  radius.     Hence  for  the  hollow  shaft 


and  for  the  solid  shaft 

•L  -  ?*L  - 

c     ~  i6d^  ~  4 

Therefore,  dividing  the  first  of  these  by  the  second,  and  letting 
k  denote  the  value  of  D/d,  there  results 
hollow  _        P+i 
solid     ~  kJ~P~-~\ 

which  is  the  ratio  of  the  strength  of  a  hollow  shaft  to  a  solid 
one  of  the  same  sectional  area. 

The  stiffness  of  a  shaft  under  torsion  is  measured  by  the 
twisting  moment  it  can  carry  with  a  given  angle  of  torsion. 
As  seen  in  Art.  66,  this  angle  is 

SJ_      Ppl_ 

-     -- 


and  hence  the  stiffness  varies  directly  as  the  polar  moment  of 
inertia.     For  the  hollow  shaft 


/  =  A*(/>  '  -  d  ')  =  \A(D  '  +  </'), 
and  for  the  solid  shaft 


Therefore,  dividing  the  first  by  the  second,  and  designating 
the  quantity  D/d  by  k, 

hollow       k*  +  i 

solid     ~  £'  -  i1 

which  is  the  ratio  of  the  stiffness  of  a  hollow  shaft  to  a  solid 
one  of  the  same  sectional  area. 

The  resilience  of  a  shaft  is  measured  by  the  work  required  to 
produce  a  given  unit-stress.     From  (26)  of  the  last  Article  this 


2$O  SHEAR   AND   TORSION.  CH.  XII. 

is  seen  to  vary  with  J/c*.     Thus  for  the  hollow  shaft 


and  for  the  solid  shaft 


Dividing  the  first  by  the  second,  or  putting  D/d  equal  to  k, 
hollow        k*  +  i 
solid  k*     ' 

which  is  the  ratio  of  the  resilience  of  a  hollow  shaft  to  a  solid 
one  of  the  same  sectional  area. 

In  practice  the  outer  diameter  is  often  about  twice  the  inner 
diameter.  For  this  case  k  =  2,  and  the  above  formulas  show 
that  a  hollow  shaft  has  1.44  times  the  strength,  1.67  times  the 
stiffness,  and  1.2$  times  the  resilience  of  a  solid  shaft  of  the 
same  sectional  area.  These  conclusions  are  valid  only  when 
the  maximum  stress  is  within  the  elastic  limit  of  the  material ; 
for  higher  stresses  a  theoretic  comparison  cannot  be  satisfac- 
torily made. 

Shafts  of  the  same  material  and  length  are  of  the  same 
strength  when  their  values  of  J/c  are  equal,  of  the  same  stiff- 
ness when  their  values  of  /are  equal,  and  of  the  same  resili- 
ence when  their  values  of  J/c*  are  equal.  It  is  thus  easy  to 
show  that  the  percentage  of  weight  saved  by  using  a  hollow 
shaft  instead  of  a  solid  one  is 

For  equal  strength,  100 

For  equal  stiffness, 

\ 

100 
For  equal  resilience,  , ,   , — , 

K     -f-    I 

ci  which  k  is  the  ratio  of  the  outer  to  the  inner  diameter  of 


ART.  124.  SHAFT  COUPLINGS.  28 1 

the  hollow  shaft.  See  a  paper  by  R.  W.  DAVENPORT  in  the 
Transactions  of  the  Society  of  Naval  Architects  and  Marine 
Engineers  for  1893  (reprinted  in  Engineering  News,  Nov.  23, 
1893)  for  a  discussion  of  the  practical  advantages  of  hollow 
shafts  made  of  steel  having  a  high  elastic  limit. 

Prob.  172.  Compare  the  strength  of  a  solid  shaft  13  inches 
in  diameter  with  that  of  a  hollow  shaft  with  outer  diameter  17 
inches  and  inner  diameter  1 1  inches,  the  elastic  strengths  being 
30000  and  50000  pounds  per  square  inch  respectively. 

ART.  124.    SHAFT  COUPLINGS 

At  A  and  B  in  Fig.  70  are  shown  the  end  and  side  views  of 
a  flange  coupling  for  a  shaft,  the  flanges  being  connected  by 
bolts.  These  bolts  in  transmitting  the  torsion  from  one  flange 
to  another  are  subject  to  shearing  stress,  and  they  must  be 


B  CD 

FIG.  70. 

of  sufficient  strength  to  safely  carry  it.  This  shear  differs  from 
that  in  the  main  body  of  the  shaft  only  in  intensity,  and  it  is 
the  greatest  upon  the  side  of  the  bolt  most  remote  from  the 
axis. 

Let  /  be  the  polar  moment  of  inertia  of  the  cross-section  of 
a  solid  shaft,  c  its  radius,  and  St  the  shearing  unit-stress  on  the 
outer  surface.  Let  Jl  be  the  polar  moment  of  inertia  of  the 
cross-section  of  the  bolts,  cl  the  distance  from  the  axis  of  the 
shaft  to  the  side  of  the  bolts  farthest  from  the  axis,  and  let  the 
shearing  unit-stress  on  that  side  be  the  same  as  that  on  the 
outer  surface  of  the  shaft.  Then  in  order  that  the  bolts  may 
be  equal  in  strength  to  the  shaft  it  is  necessary  that  J/c  should 
equal  JJcv  The  polar  moment  of  inertia  of  the  cross-section 
of  one  bolt  with  respect  to  the  axis  of  the  shaft  is  equal  to  its 


282  SHEAR  AND  TORSION.  CH.  XII. 

polar  moment  with  respect  to  its  own  axis  plus  the  area  of  the 
cross-section  into  the  square  of  the  distance  between  the 
two  axes. 

Let  D  be  the  diameter  of  the  shaft,  d  the  diameter  of  each 
of  the  bolts,  h  the  distance  of  the  center  of  a  bolt  from  the  axis 
of  the  shaft,  and  n  the  number  of  bolts  ;  then 


and  equating  these  values  there  is  found 

D  \d  +  2h)  =  nd\d*  +  Stf), 

which  is  the  necessary  relation  between  the  quantities  in  order 
that  the  bolts  may  be  equal  in  strength  to  the  shaft,  provided 
the  material  be  the  same. 

This  formula  is  an  awkward  one  for  determining  d,  and  hence 
it  is  often  assumed  that  the  shear  is  uniformly  distributed  over 
the  bolts,  or  that  ^  =  h  and  /,  =  ^nd^h  a.  This  amounts  to 
the  same  thing  as  regarding  d  as  small  compared  to  h,  and 
the  expression  then  reduces  to 

,         or        d= 

In  practice  the  bolts  are  often  made  a  little  larger  in  diameter 
than  this  formula  requires. 

The  above  supposes  the  shaft  to  be  solid.  If  it  be  hollow 
with  outer  diameter  D  and  inner  diameter  dl  ,  the  D  3  in  the 
above  expressions  is  to  be  replaced  by  Z>*  —d*/D,  if  dl  be 
large  enough  to  have  any  influence. 

The  case  shown  at  CD  in  Fig.  70  is  one  that  would  not  occur 
in  practice,  but  it  is  here  introduced  in  order  to  indicate  that 
the  bolts  would  be  subject  to  a  flexural  as  well  as  a  shearing 
stress.  It  is  clear  that  the  flexural  stress  will  increase  with  the 
length  of  the  bolts,  and  that  they  should  be  greater  in  diameter 
than  for  the  case  of  pure  shearing.  The  flexural  stress  will 


ART.  125.  A  CRANK   PIN  AND   SHAFT.  283 

also  depend  upon  the  work  transmitted  by  the  shaft.  This 
case  will  be  investigated  in  Art.  126  in  connection  with  the  dis- 
cussion of  the  pin  of  a  crank  shaft. 

Prob.  173.  A  solid  shaft  6  inches  in  diameter  is  coupled  by 
bolts  li  inches  in  diameter  with  their  centers  5  inches  from  the 
axis.  How  many  bolts  are  necessary  ? 

Prob.  174.  A  hollow  shaft  17  inches  in  outer  and  II  inches 
in  inner  diameter  is  to  be  coupled  by  12  bolts  placed  with  their 
centers  20  inches  from  the  axis.  What  should  be  the  diameter 
of  the  bolts  ? 

ART.  125.    A  CRANK  PIN  AND  SHAFT. 

A  crank  pin,  CD  in  Fig.  71,  is  subject  to  a  pressure  W  from 
the  connecting  rod  which  is  uniformly  distributed  over  nearly 
its  entire  length.  This  pressure  varies  at  different  positions  in 
the  stroke  of  the  engine,  but  for  ordinary  computations  may 
be  taken  at  from  10  to  20  per  cent  greater  than  the  total  mean 
pressure  on  the  steam  piston ;  to  this  may  be  added  the  weight 
of  the  connecting  and  piston  rods  in  case  these  should  be  ver- 
tical in  position. 

This  maximum  pressure  W  causes  a  cross-shear  in  the  crank 
pin  at  the  section  C,  and  it  also  causes  a  flexural  stress  at  C 
due  to  a  uniform  load  over  the  cantilever 

CD.     These  may  be   computed    by  the      

methods  of  Chapter  III,  and  their  com-  {  A 
bined  influence  can  be  determined  by 
formula  (13)  of  Art.  75.  The  compressive 
or  bearing  stress  on  the  surface  of  the  pin 
is  usually  also  to  be  considered,  this  being 
estimated  per  square  unit  of  diametral  j 

area. 

Owing  to  the  constant  alternation  of  these  stresses  as  the 
crank  arm  revolves,  the  allowable  working  unit-stresses  should 


\w 


284 


SHEAR  AND  TORSION. 


CH.  XII. 


be  taken  low  in  designing  the  pin,  arm,  and  journal  bearing. 
The  crank  arm  is  under  flexure  as  a  cantilever  loaded  at  the 
end,  while  the  part  of  the  shaft  AB  which  rests  in  the  journal 
is  subject  to  combined  shearing,  flexure,  and  torsion.  The 
methods  of  Chapters  VI  and  VII  give  all  required  to  thor- 
oughly discuss  these  cases.  See  UNWIN'S  Elements  of  Machine 
Design  for  the  special  formulas  generally  used  in  practice. 

Prob.  175.  The  crank  CD  in  Fig.  71  is  8  inches  long  and  4 
inches  in  diameter,  the  maximum  pressure  W  being  60000 
pounds.  Compute  the  bearing  unit-stresses,  the  shearing  unit  - 
stress  and  the  flexural  unit-stress.  Compute  the  maximum 
unit-stress  due  to  combined  shear  and  flexure. 

ART.  126.    A  TRIFLE-CRANK  PIN. 

Double  and  triple  cranks  are  used  when  several  engines  are 
to  be  attached  to  the  same  shaft,  as  is  usual  in  ocean  steamers. 
With  the  triple  arrangement  the  cranks  are  set  at  angles  of 
1 20  degrees  with  each  other,  thus  securing  a  uniform  action 
upon  the  shaft.  Fig.  72  shows  one  of  these  cranks,  AB  and 


B 
C 

k 

# 

XT' 

C3 

F{ 

c° 

^ 

D 

[  (J 

i                       vL 

FIG.  72. 

EF  being  portions  of  the  shaft  resting  in  journal  bearings,  CD 
the  crank  pin  to  which  the  connecting  rod  is  attached,  while 
BC  and  DE  are  the  crank  arms  or  webs  which  are  usually 
shrunk  upon  the  shaft  and  pins. 

The  complete  investigation  of  the  maximum  stresses  in  such 
a  crank    shaft  and  pin  is  one    of   much   difficulty.      A  brief 


ART.   126.  A  TRIPLE-CRANK   PIN.  28$ 

abstract  of  such  an  investigation  will,  however,  here  be  given 
for  the  crank  pin.  There  are  three  cranks,  and  the  one  to  be 
considered  is  the  nearest  to  the  propeller,  so  that  the  torsion 
from  the  other  two  cranks  is  transmitted  through  the  pin  CD. 
This  steel  crank  pin  is  hollow,  18  inches  in  outer  diameter  and 
6  inches  in  inner  diameter,  its  length  between  webs  being  24 
inches,  the  thickness  of  each  web  12  inches,  and  the  distance 
from  the  axis  of  the  shaft  to  the  center  of  the  pin  being  30 
inches.  The  three  engines  transmit  7200  horse-power  to  the 
shaft  EF,  of  which  4800  horse-power  is  transmitted  through  the 
shaft  AB  and  through  the  crank  pin  CD.  The  maximum 
pressure  W  brought  by  the  connecting  rod  upon  the  crank  pin 
is  156000  pounds.  It  is  required  to  determine  the  stresses 
when  the  crank  makes  80  revolutions  per  minute. 

The  pressure  W  is  distributed  over  about  17  inches  of  the 
length  of  the  pin,  so  that  the  bearing  compressive  stress  on  the 
diametral  area  is 

156000 
5'  ==  YffiS  _  6)  =  765  pounds  per  square  inch, 

which  is  a  low  and  safe  value. 

The  shearing  stress  due  to  W,  taken  as  uniformly  distributed 
over  the  cross-section  of  the  pin,  is 

_  78000 

5' =  oTs^cTs'"^3)  =  345  pounds  per  S(*uare  mch' 

which  is  low,  but  will  be  much  increased  by  the  other  stresses 
acting  on  the  end  section. 

The  shearing  stress  due  to  the  horse-power  transmitted 
through  BC  has  its  greatest  value  on  the  side  of  the  pin 
farthest  from  the  axis.  The  twisting  moment  Pp  due  this 
4800  horse-power  is  found,  from  the  first  equation  on  page  141 
of  Art.  67,  to  be 

198000  X  4800 
Pp  =      3.1416  X  80     =  3  782  ooo  pound-inches, 


286  SHEAR  AND   TORSION.  CH.  XII. 

and  this  is  equal  to  the  resisting  moment  of  the  crank  pin,  or 
to  SJJcl ,  in  which  Jl  is  the  polar  moment  of  inertia  of  the 
cross-section  with  respect  to  the  axis  of  the  shaft  and  cl  is  the 
distance  from  that  axis  to  the  side  of  the  pin  where  the  stress 
6"  is  to  be  found.  Now,  c  =  30  +  9  =  39  inches,  and  then 
/,  =  0.0982(18*  —  6')  +  0.7854(18'  -  6')  X  39'.  Hence,  from 
formula  (i  i)  of  Art.  64, 

Pp,c,       3  782  ooo  X  39 

o,  =  —=-—  = -  =  420  pounds  per  square   inch, 

/i  354000 

which  is  the  maximum  shearing  stress  due  to  the  transmitted 
power. 

The  flexure  of  the  pin  due  to  the  torsion  carried  through  it 
falls  under  a  case  not  heretofore  considered,  except  in  the  brief 
mention  in  Art.  124.  The  twisting  moment  Pp  is  equivalent 
to  a  force  P  acting  at  a  distance  of  30  inches  from  the  shaft 
and  normal  to  the  crank  arms  ;  the  value  of  P  is 

3  782000 
P  =  — =126  100  pounds, 

and  this  produces  a  bending  moment  in  the  pin  which  may  be 
taken  as  a  beam  fixed  at  both  ends  while  P  acts  in  opposite 
directions  at  those  ends.  Hence  there  is  a  bending  moment 
M'  at  each  end,  opposite  in  sign  but  equal  in  value,  and  the 
moment  at  any  section  is  M  •=.  M'  -f-  Px ;  but  when  x  =  /the 
value  of  M  is  —  M'  and  therefore  M'  =  ±  J/V,  which  is  the 
maximum  bending  moment.  Thus  from  (4)  of  Art.  21 

.  M'c  _  63000  X  18  X  9  _       .. 
**-      I  ^,r(i84-64) 

which  is  the  flexural  stress  in  pounds  per  square  inch. 

All  of  these  stresses  are  light,  but  the  pin  is  necessarily  made 
heavier  than  they  would  require  on  account  of  the  additional 
stresses  due  to  the  shrinking  of  the  web  upon  the  pin.  The 
data  here  given  are  not  sufficient  to  determine  these  with  ex- 
actness, but  there  is  a  radial  compressive  unit-stress  56  brought 


ART.  126.  A   TRIPLE-CRANK   PIN.  287 

by  the  web  upon  the  pin  of  probably  3  ooo  pounds  per  square 
inch,  and  this  is  accompanied  by  a  tangential  compressive  unit- 
stress  S6  of  about  4000  pounds  per  square  inch  (Art.  142). 
These  take  effect  in  the  fillet  of  the  pin  on  the  inside  of  the 
web  at  D,  where  also  all  of  the  other  stresses  concentrate 
except  5,.  In  Art.  134  it  will  be  shown  how  these  several 
values  may  be  combined  in  order  to  obtain  the  maximum 
tensile,  compressive,  and  shearing  stresses. 

Prob.  176.    Draw   the  shear  and   moment  diagram  for  the 
lateral  flexure  of  the  pin  due  to  the  transmitted  torsion. 


288  APPARENT  AND  TRUE  STRESSES.  CH.  XIII. 


CHAPTER  XIII. 
APPARENT  STRESSES  AND  TRUE  STRESSES. 

ART.  127.    THE  MATHEMATICAL  THEORY  OF  ELASTICITY. 

In  Art.  5  are  stated  several  laws,  derived  from  experiment, 
which  are  the  foundation  of  the  science  of  Mechanics  of  Ma- 
terials. Of  these  (A)  and  (B)  relate  to  elasticity  and  are  the 
basis  of  all  discussions  concerning  stresses  that  do  not  surpass 
the  elastic  strength  of  the  material,  the  latter  being  usually  re- 
ferred to  as  HOOKE'S  law  (Art.  81).  All  the  theoretic  formulas 
of  the  preceding  pages  are  derived  by  the  help  of  this  law,  and 
these  hence  constitute  a  part  of  the  mathematical  theory  of 
elasticity. 

This  theory  is  one  of  vast  extent  and  far-reaching  conse- 
quences, and  its  full  development  would  require  volumes.  It 
includes  not  only  the  complete  investigation  of  the  stresses 
and  deformations  produced  in  every  part  of  a  body  by  gradu- 
ally applied  exterior  forces,  but  also  those  arising  under  condi- 
tions of  impact.  It  deals  not  only  with  elastic  solids,  but  with 
fluids,  gases,  and  the  ether  of  space.  The  discussion  of 
stresses  and  deformations,  both  in  homogeneous  and  crystalline 
bodies,  leads  to  the  investigation  of  wave  propagations,  the 
time  and  velocity  of  elastic  oscillations,  and  numerous  other 
phenomena  of  physics.  In  this  Chapter  will  be  presented  a 
few  of  the  fundamental  principles  with  reference  to  homoge- 
neous materials  only. 

Statics  proper  is  concerned  only  with  rigid  bodies,  while  the 
theory  of  elasticity  deals  with  bodies  deformed  under  the  action 
of  exterior  forces  and  which  recover  their  original  shape  on  the 


ART.  127.    MATHEMATICAL  THEORY   OF   ELASTICITY.  289 

removal  of  these  forces.  All  the  principles  and  methods  of 
statics  apply  in  the  discussion  of  elastic  bodies,  but  in  addition 
new  principles  based  upon  HOOKE'S  law  arise.  The  amount 
of  deformation  being  small  within  the  elastic  limit  for  common 
materials,  it  is  allowable  to  neglect  the  squares  and  higher 
powers  of  a  unit-elongation  in  comparison  with  the  elongation 
itself.  Thus  if  /  be  the  length  of  a  bar,  which  under  the  action 
of  stress  is  increased  to  the  length  l(\  +  s\  the  square  of  this 
new  length  may  be  taken  as  l\i  -\-  2s)  and  the  cube  as 
l\i  +  3*).  For  a  substance  like  india  rubber,  where  this  as- 
sumption does  not  apply,  some  of  the  conclusions  of  the  theory 
of  elasticity  are  not  necessarily  valid. 

Another  axiom  derived  from  experience  is  the  following: 
Under  tensile  forces  the  volume  of  a  body  is  increased  and 
under  compressive  forces  it  is  diminished.  This  is  the  case  for 
the  common  materials,  although  bodies  may  exist  for  which  it 
is  not  true.  Thus  if  a  cube  be  compressed  upon  two  opposite 
faces  the  edges  parallel  to  the  forces  are  decreased  while  those 
at  right  angles  to  the  forces  are  increased  in  length  ;  on  the 
whole,  however,  the  volume  of  the  cube  is  slightly  decreased. 

The  student  should  consult  the  article  on  Elasticity  by 
KELVIN  in  the  Encyclopaedia  Britannica,  as  also  the  History 
of  TODHUNTER  and  PlERSON.  The  works  of  CLEBSCH  (Elas- 
ticitat fester  Korper,  1862),  WlNKLER  (Elasticitat  und  Festig- 
keit,  1867),  GRASHOF  (Theorie  der  Elasticitat  und  Festigkeit, 
1878),  and  FLAMANT  (Resistance  des  Materiaux,  1886)  maybe 
mentioned  as  treating  the  subject  both  from  the  theoretical 
and  the  engineering  point  of  view. 

Prob.  177.  Consult  TODHUNTER  and  PlERSON'S  History  of 
the  Theory  of  Elasticity  and  of  the  Strength  of  Materials, 
and  ascertain  something  about  the  investigations  of  SAINT 

VENANT. 


290  APPARENT  AND  TRUE  STRESSES.        CH.  XIII. 

ART.  128.    LATERAL  DEFORMATION. 

It  has  already  been  noted,  in  Arts.  71  and  121,  that  a  bar 
under  tension  not  only  elongates  in  the  direction  of  its  length, 
but  is  subject  to  a  lateral  contraction.  So  if  a  bar  be  under 
compression  there  occurs  a  longitudinal  contraction  and  a 
lateral  elongation.  If  P  be  the  applied  force  and  A  the  area 
of  the  cross-section  the  unit-stress  is  P/A  =5;  if  /  be  the 
length. and  A  the  longitudinal  change  of  length,  then  A//  =  s  is 
the  unit-elongation  or  unit-contraction.  In  the  case  of  tension 
each  unit-length  of  the  bar  is  increased  by  s,  but  each  unit  at 
right  angles  to  the  length  is  decreased  by  the  amount  es,  where 
e  is  an  abstract  number  less  than  unity,  called  the  factor  of 
lateral  contraction.  In  the  case  of  compression  es  is  the  lateral 
unit-elongation. 

Let  a  cube  in  the  interior  of  a  bar  under  tension  have  each 
of  its  sides  unity  before  the  application  of  the  stress ;  under  this 
tension  the  cube  is  deformed  so  that  its  sides  are  i  -{-s,  l  —  es, 
and  I  —  es.  The  volume  of  the  deformed  body  is,  if  the 
squares  and  cubes  of  s  be  neglected  in  comparison  with  the 
first  power, 

(l+*Xl-eO'   =    I  +(1-26)5. 

Now  in  order  that  the  volume  may  be  increased,  (i  —  2e)s 
must  be  positive,  or  e  must  be  less  than  J.  When  e  is  o  no 
lateral  contraction  occurs,  when  e  =  £  the  lateral  contraction  is 
a  maximum.  Thus  for  all  bodies  whose  volume  is  increased 
under  tension  the  factor  of  lateral  contraction  must  lie  be 
tween  o  and  J. 

The  same  reasoning  applies  in  the  case  of  compression,  for 
which  as  far  as  known  e  has  the  same  value  as  for  tension.  By 
precise  measurements  of  bars  under  stresses  within  the  elastic 
limit  it  has  been  found  that  e  generally  lies  between  0.2  and 
0.4  for  wrought  iron  and  steel,  a  mean  value  extensively  used 
being  -J.  For  cast  iron  e  is  about  J  or  a  little  less. 


ART.  129.   TRUE  TENSILE  AND   COMPRESSIVE  STRESSES.      291 

Prob.  178.  A  bar  of  steel  2X2  inches  and  6  feet  long  is 
pulled  by  a  force  of  50  ooo  pounds.  Compute  the  percentage 
of  increase  of  length,  and  the  percentage  of  increase  of  volume. 

ART.  129.    TRUE  TENSILE  AND  COMPRESSIVE  STRESSES. 

If  a  parallelepiped  be  subject  to  a  unit-stress  5  in  the  direc- 
tion of  its  length  this  is  the  true  stress  on  all  planes  normal  to 
its  length.  In  directions  at  right  angles  to  the  length  there 
exists,  however,  a  lateral  contraction  which  implies  an  internal 
stress  of  compression.  If  s  be  the  unit-elongation  due  to  S, 
the  lateral  unit-contraction  is  es,  and  this  is  the  same  as  would 
be  produced  by  a  lateral  compressive  unit-stress  eS.  Thus  it 
is  clear  that  the  deformation  at  right  angles  to  5  is  the  same  as 
that  produced  by  an  actual  unit-stress  eS.  In  a  similar  manner 
if  forces  act  upon  all  the  sides  of  the  parallelepiped  the  true 
internal  stresses  are  different  from  the  apparent  ones. 

All  the  stresses  computed  thus  far  in  this  volume  are  appar- 
ent stresses,  for  the  influence  of  lateral  deformation  has  not 
been  taken  into  account.  In  this  Chapter  the  letter  5  will 
denote  the  apparent  unit-stresses,  while  the  true  unit-stresses 
corresponding  to  the  actual  deformations  will  be  designated 
by  T.  The  true  resistance  of  a  body  depends  upon  the  actual 
deformations  produced,  and  these  are  measured  by  the  true 
internal  stresses. 

Let  a  homogeneous  parallelopiped  be  subject  to  normal 
forces  of  tension  or  compression 
upon  its  six  faces,  those  upon  oppo- 
site faces  being  equal.  Let  the 
edges  of  the  parallelopiped  be  de- 
signated by  01,  02,  03,  as  in  Fig.  73. 
Let  St  be  the  normal  unit-stress 
upon  the  two  faces  perpendicular  to 
the  edge  01,  and  5,  and  St  those  upon  the  faces  normal  to  02 


APPARENT  AND  TRUE  STRESSES.  CH.  XIII. 

and  03  ;  thus  the  directions  of  Sl  9  Sa,  5,  are  parallel  to  01, 
02,  03,  respectively.  Then,  supposing  that  these  stresses  are 
all  tensile,  and  that  the  factor  of  lateral  contraction  e  is  the 
same  in  all  directions,  the  true  internal  unit-stresses  in  the 
three  directions  are, 

T,  =  S,  -  eS,  -  eS, , 
(27)  r,  =  Sf-  eS,  -  eSl9 

T,  =  S>   —  eS,  —  e-S, , 

in  which  e  lies  between  o  and  £,  as  shown  in  Art.  128.  If  any 
apparent  stress  5  be  compressive,  it  is  to  be  taken  as  negative 
in  the  formulas,  and  then  the  true  stresses  7",,  J!,,  T9  are 
tensile  or  compressive,  according  as  their  numerical  values  are 
positive  or  negative. 

For  example,  let  a  cube  be  stressed  upon  all  sides  by  the 
apparent  compression  5 ;  then  the  true  internal  unit-stress  T 
is  S(i  —  2e),  or  about  £5,  and  its  linear  deformation  is  only 
about  one-third  of  that  due  to  a  compressive  stress  5  applied 
upon  two  opposite  faces.  Again,  if  a  bar  have  a  tension  S^in 
the  direction  of  its  length,  and  no  apparent  stresses  upon  its 
sides,  then  Tl  =  5,  while  7",  =  Tt  =  —  eS,. 

The  true  deformations  corresponding  to  the  true  internaV 
stresses  will  be  denoted  by  ^  ,  /,,  /,.  If  the  coefficient  of 
elasticity  in  all  directions  be  £,  then 

T  T  T 

/_±j        /_!_«        /____» 

**        E'        '        £'        '  ""   E 

\ 

are  the  unit-elongations  or  unit-contractions  parallel  to  the 
three  coordinate  axes. 

As  a  simple  example,  let  a  steel  bar  2  feet  long  and  3X2 
inches  in  cross-section  be  subject  to  a  tension  of  60000  pounds 
in  the  direction  of  its  length  and  to  a  compression  of  432  ooo 
pounds  upon  the  two  opposite  flat  sides.  Here  Sl  =  60  000/6 
•=  10  ooo  pounds  per  square  inch,  5a  =  —  432  000/72  =  —  6000 


ART.  130.      NORMAL  AND  TANGENTIAL   STRESSES.  293 

pounds  per  square  inch,  and  .S,  =  o.     Then  from  (27),  taking 
€  as      the  true  internal  stresses  are 


7\  =  +12  ooo,  7;  =  -9330,  r,  = 
and  it  is  thus  seen  that  the  true  tensile  unit-stress  is  20  per 
cent  greater  than  the  apparent,  while  the  true  compressive 
unit-stress  is  more  than  50  per  cent  greater  than  the  apparent. 

If  the  parallelepiped  in  Fig.  73  be  subject  to  the  action  of 
oblique  stresses  Rt  ,  Rt  ,  Rt  ,  each  may  be  resolved  into  a  stress 
normal  to  the  face  and  into  shearing  stresses  parallel  to  the 
edges.  In  such  a  case  the  true  unit-stresses  Tl  ,  7",  ,  7",  can- 
not be  directly  found,  but  it  will  be  shown  in  the  following 
articles  that  three  planes  can  be  determined  upon  which  the 
apparent  stresses  are  wholly  normal,  and  that  these  are  the 
maximum  apparent  tensile  and  compressive  stresses  due  to 
Rt,  RtJ  R%\  these  being  found,  formulas  (27)  are  directly 
applicable. 

Prob.  179.  In  Fig.  73  let  a  plane  be  passed  through  the  edge 
02  and  through  the  edge  diagonally  opposite  to  it.  Let  the 
edges  01,  02,  03  be  equal  in  length.  Show  that  the  apparent 
shearing  unit-stress  on  this  plane  is  J(5,  —  S,). 


ART.  130.    NORMAL  AND  TANGENTIAL  STRESSES. 

The  general  case  of  internal  stress  is  that  of  an  elementary 
parallelepiped  held  in  equilibrium  by  apparent  stresses  applied 
to  its  forces  in  directions  not  normal.  Here  each  oblique 
stress  may  be  decomposed  into  three  components  parallel  re- 
spectively to  three  coordinate  axes,  OX,  O  Y,  OZ.  Upon  each 
of  the  faces  perpendicular  to  OX  the  normal  component  of  the 
oblique  unit-stress  is  designated  by  Sx  and  the  two  tangential 
components  by  S^  and  Sxx.  A  similar  notation  applies  to  each 
of  the  other  faces.  An  5  having  but  one  subscript  denotes  a 
tensile  or  compressive  stress,  and  its  direction  is  parallel  to  the 


294 


APPARENT  AND  TRUE   STRESSES.  CH.  XIIL 


axis  corresponding  to  that  subscript.  An  5  having  two  sub- 
scripts denotes  a  shearing  stress,  the  first  subscript  designating 
the  axis  to  which  the  face  is  perpendicular  and  the  second 
designating  the  axis  to  which  the  stress  is  parallel ;  thus  Sgx  is 


FIG.  74. 

on  the  face  perpendicular  to  OZ  and  its  direction  is  parallel  to 
OX.  In  Fig.  74  the  six  components  for  three  sides  of  the 
parallelepiped  are  shown.  Neglecting  the  weight  of  the  par- 
allelepiped the  components  upon  the  three  opposite  sides  must 
be  of  equal  intensity  in  order  that  equilibrium  may  obtain. 

An  elementary  parallelepiped  in  the  interior  of  a  body  is 
thus  held  in  equilibrium  under  the  action  of  six  normal  and 
twelve  tangential  stresses  acting  upon  its  faces.  The  normal 
stresses  upon  any  two  opposite  faces  must  be  equal  in  intensity 
and  opposite  in  direction.  The  tangential  stresses  upon  any 
two  opposite  faces  must  also  be  equal  in  intensity  and  opposite 
in  direction. 

A  certain  relation  must  also  exist  between  the  six  shearing 
stresses  shown  in  Fig.  74  in  order  that  equilibrium  may  obtain. 
Let  the  parallelepiped  be  a  cube  with  each  edge  equal  to  unity ; 
then  if  no  tendency  to  rotation  exists  with  respect  to  an  axis 
through  the  center  of  the  cube  and  parallel  to  OX  it  is  neces- 
sary that  Syz  should  equal  Sgy.  A  similar  condition  obtains  for 
each  of  the  other  rectangular  axes,  and  hence 

(<>Q\  C      _    C  C      _    C  C      _    C 

\    ^/  *^xy  ~~~   *-^yx  t  *-^yz  ~~~   *^xyi  ^xz  —  *^sx  t 


ART.   131.  RESULTANT   STRESS.  295 

that  is,  those  shearing  unit-stresses  are  equal  which  are  upon 
any  two  adjacent  faces  and  normal  to  their  common  edge. 

The  apparent  unit-stresses  designated  by  5  are  computed  by 
the  methods  of  the  preceding  Chapters  ;  it  is  rare,  however, 
that  more  than  three  or  four  of  them  exist,  even  under  the  action 
of  complex  forces.  The  general  problem  is  then  to  find  a 
parallelepiped  such  that  the  resultant  stresses  upon  it  are 
wholly  normal.  These  resultant  normal  stresses  will  be  5, ,  .5, , 
S9,  from  which  by  (27)  the  true  normal  stresses  Tt,  7,.  T9 
can  be  found.  It  will  later  be  shown  that  these  stresses  5, ,  5, , 
5",  are  the  maximum  apparent  stresses  of  tension  or  compres- 
sion resulting  from  the  given  normal  and  tangential  stresses. 

Prob.  1 80.  Let  a,  b,  c  be  the  angles  which  a  line  makes  with 
the  axes  OX,  OY,  OZ,  respectively.  Show  that  the  sum  of 
the  squares  of  the  cosines  of  these  angles  is  equal  to  unity. 

ART.  131.    RESULTANT  STRESS. 

The  resultant  unit-stress  upon  any  face  of  the  parallelepiped 
in  Fig.  74  is  the  resultant  of  the  three  rectangular  unit-stresses 
acting  upon  that  face.  Thus  for  the  face  normal  to  OZ  the 
resultant  unit-stress  is  given  by 

R:  =  S:-\-S'^  +  S\,, 

and  the  total  resultant  stress  upon  that  face  is  the  product  of 
its  area  and  R9 . 

The  resultant  unit-stress  R  upon  any  elementary  plane  hav- 
ing any  position  can  be  determined  when  the  normal  and  tan- 
gential stresses  in  the  directions  parallel  to  the  coordinate  axes 
are  known.  Let  a  plane  be  passed  through  the  corners  I,  2,  3, 
of  the  parallelepiped  in  Fig.  74,  and  let  a,  b,  c  be  the  angles 
that  its  normal  makes  with  the  axes  OX,  O  Y,  OZ,  respectively. 
Let  a,  ft,  y  be  the  angles  which  the  resultant  unit-stress  R 
makes  with  the  same  axes.  Let  A  be  the  area  of  the  triangle 


296  APPARENT  AND  TRUE  STRESSES.  CH.  XIII. 

123;  then  the  total  resultant  stress  upon  that  area  is  AR,  and 
its  components  parallel  to  the  three  axes  are  AR  cos  a,  AR 
cos  ft,  AR  cos  y.  The  triangle  whose  area  is  A,  together  with 
the  three  triangles  012,  023,  031,  form  a  pyramid  which  is  in 
equilibrium  under  the  action  of  R  and  the  stresses  upon  the 
three  triangles.  The  areas  of  these  triangles  are  A  cos  a, 
A  cos  b,  A  cos  c,  and  the  stresses  upon  them  are  the  products  of 
the  areas  by  the  several  unit-stresses.  Now  the  components  of 
these  four  stresses  with  respect  to  each  rectangular  axis  must 
vanish  as  a  necessary  condition  of  equilibrium.  Hence,  can- 
celling out  A,  which  occurs  in  all  terms,  there  results 

R  cos  a  =  S*  cos  a  -f-  Syjf  cos  b  -f-  Szx  cos  c, 
(29)  R  cos  ft  =  S^y  cos  a  -\-  Sy  cos  b  -\-  S,y  cos  c, 
R  cos  Y  =  Sxz  cos  a  -f-  Sy,  cos  b  -\-  Ss  cos  c, 

in  which  the  second  members  are  all  known  quantities. 

From  these  equations  the  values  of  R  cos  <*,  R  cos  ft,  R  cos  y 
can  be  computed ;  then  the  sum  of  the  squares  of  these  is  R* 
since  cos*  a  -f-  cosa  ft-\-  cos3  y  =  i.  The  value  of  cos  a  is  found 
by  dividing  that  of  R  cos  a  by  R,  and  similarly  for  cos  ft  and 
cos  /.  Now  the  angle  0  between  the  directions  of  R  and  the 
normal  to  the  plane  is  given  by 

cos  B  ==  cos  a  cos  a  +  cos  b  cos  /?  -|-  cos  c  cos  ^, 

and  then  the  tensile  or  compressive  unit-stress  normal  to  the 
given  plane  is  R  cos  0,  while  the  resultant  shearing  unit-stress 
is  R  sin  8.  This  shearing  stress  may  be  resolved  into  two  com- 
ponents in  any  two  directions  on  the  plane. 

As  a  simple  numerical  example,  let  a  bolt  be  subject  to  a 
tension  of  12000  pounds  per  square  inch  and  also  to  a  cross- 
shear  of  8  ooo  pounds  per  square  inch.  It  is  required  to  find 
the  apparent  unit-stresses  on  a  plane  making  an  angle  of  60 
degrees  with  the  axis  of  the  bolt.  Take  OX  parallel  to  the 
tensile  force  and  OY  parallel  to  the  cross-shear.  Then  Sx  = 


ART.  132.  THE   ELLIPSOID   OF  STRESS.  297 

-f  12000,  Sjej,  =  8000,  Sy*  =  8000,  and  the  other  stresses  are 
zero  ;  also  a  =  30°,  b  =  60°,  and  c  =  90°.     Then  from  (29) 
R  cos  a  =  +  14  390,  R  cos  /?  =  -f-  6930,  7?  cos  y  =  o, 
and  the  resultant  stress  in  the  plane  is, 

R  =  -/I4  390'  +  693°'  =  15  970  pounds  per  square  inch. 
The  direction  made  by  R  with  the  axis  is, 


cos  a  =  -_  =  0.901,        a  = 


cos  ft  =  =  0.434,        ft  =  25t°, 

and  the  angle  between  the  resultant  R  and  the  normal  to  the 
plane  is  given  by 

cos  6  =  0.866  X  0.901  +  0.5  X  0.434  =  0.997. 
Lastly,  the  normal  tensile  stress  on  the  plane  is  found  to  be 
R  cos  0=  15  920  pounds  per  square  inch,  while  the  shearing 
stress  on  the  plane  is  R  sin  6  =  1200  pounds  per  square  inch. 

Prob.  1  8  1.  Find  for  the  above  example  the  position  of  a 
plane  upon  which  there  is  no  shearing  stress. 

ART.  132.    THE  ELLIPSOID  OF  STRESS. 

The  resultant  unit-stress  R  upon  any  plane  makes  an  angle 
B  with  the  normal  to  that  plane.  If  at  any  point  this  plane 
be  supposed  to  vary  its  direction  the  intensity  of  R  will  be 
represented  by  the  radius  vector  of  an  ellipsoid. 

Let  Rt  ,  RI  ,  Rt  be  the  resultant  unit-stresses  upon  the  three 
faces  of  the  parallelepiped  in  Fig.  74,  and  let  0,  ,  0,  ,  0,  be  the 
angles  which  they  make  with  the  coordinate  axis  OX\  then 

cos  0,  =  J?,         cos  0,  =  §?,        cos  0,  =  %^, 
Ki  K*  <K9 

determine  the  directions  of  R,  ,  R9  ,  Rt  .  Now  let  these  direc- 
tions be  taken  as  those  of  a  new  system  of  oblique  coordinate 
axes,  let  R  be  the  resultant  unit-stress  in  any  direction,  and  let 


298 


APPARENT  AND  TRUE   STRESSES. 


CH.  XIII. 


Rx,  Ry,  Rz  be  its   components  parallel   to   these  new  axes. 
Then  R  cos  a  is  the  component  of  R  parallel  to  OX,  and 

R  cos  a  =  Rx  cos  6,  -f-  Ry  cos  #a  +  Rs  cos  #, , 
or,  inserting  for  the  cosines  their  values, 


Comparing  this  with  the  first  equation  in  (30),  it  is  seen  that, 

Rx  .      Rv  R. 

cos  a  =  -75-,         cos  b  =  •—,         cos  c  =  ^-. 

But  the  sum  of  the  squares  of  these  cosines  is  equal  to  unity ; 
therefore, 


&?        R?       R 

in  which  the  numerators  are  variable  coordinates  and  the  de- 
nominators are  given  quantities.  This  is  hence  the  equation 
of  the  surface  of  an  ellipsoid  with  respect  to  three  coordinate 
axes  having  the  directions  of  Rl ,  Rt ,  R3 . 

The  ellipsoid  is  hence  a  figure  whose  radius  vector  repre- 
sents the  resultant  unit-stress  upon  a  plane  whose  normal 
makes  an  angle  6  with  the  direction  of  that  radius  vector.  If 
the  forces  be  entirely  confined  to  one  plane  the  ellipsoid  re- 
duces to  an  ellipse. 

If  there  be  three  planes  at  right  angles  to  each  other  which 
are  subject  only  to  normal  stresses,  as  in  Fig.  75,  the  normal 
unit-stresses  Sx ,  Syt  S,  correspond 
to  R1 ,  Rt ,  ^8  in  the  above  equa- 
tion of  the  ellipsoid.     In  this  case 
S*,  Sy,  S,  are  the  three  axes  of 
the   ellipsoid.     If    now   shearing- 
stresses  be  applied  to  the  faces 
the  ellipsoid  will  be  deformed,  and 
the   three    axes   will   take    other 


FIG.  75. 


position  corresponding  to  three  planes  upon  which  no  shear- 


ART.  133.          THE  THREE   PRINCIPAL  STRESSES.  299 

ing  stresses  act.     The  stresses  corresponding  to  the  axes  of 
the  ellipsoid  are  called  principal  stresses. 

Prob.  182.  If  Sy  =  Sz  in  Fig.  75,  show  that  the  ellipsoid  be- 
comes either  a  prolate  spheroid  or  an  oblate  spheroid. 

ART.  133.    THE  THREE  PRINCIPAL  STRESSES. 

In  general  the  resultant  unit-stress  R  upon  a  given  plane 
makes  an  angle  6  with  the  normal  to  that  plane,  and  hence  can 
be  resolved  into  a  normal  stress  of  tension  or  compression  and 
into  two  tangential  shearing  stresses  (Art.  131).  It  is  evident, 
however,  that  planes  may  exist  upon  which  only  normal 
stresses  act,  so  that  0  is  zero  and  R  is  pure  tension  or  com- 
pression. In  order  to  find  these  planes  and  the  stresses  upon 
them  the  angles  or,  ft,  y  in  the  equations  (29)  are  to  be  made 
equal  to  a,  b,  c,  respectively.  Also  replacing  R  by  S,  they 

become, 

(S  —  Sr)  cos  a  =  Syx  cos  b  +  $**  cos  c, 
(S  —  Sy)  cos  b  =  Sxy  cos  a  -f-  Sty  cos  c, 
(S  —  Sg)  cos  c  =  Sxg  cos  a  +  Syg  cos  bt 

in  which  S,  cos  a,  cos  b,  cos  c  are  quantities  to  be  determined. 

The  three  angles,  however,  are  connected  by  the  necessary 

relation, 

cos'  a  +  cos1  b  +  cos1  c  =  i, 

and  hence  four  equations  exist  between  four  unknowns. 

Remembering  the  relation  between  the  shearing  stresses  ex- 
pressed in  (28),  the  solution  of  the  equations  leads  to  a  cubic 
equation  for  S,  which  is  of  the  form, 


(30)  5'  -  AS9  +  £S  -  C  =  o, 

in  which  the  values  of  the  coefficients  are, 

A  =  SJf  +  S,+  S., 

B  =  5,5,  +  SfS.  +  S.S,  -  S',,  -  5% 

a      - 

yt 


300  APPARENT  AND  TRUF   STRESSES.  CH.  XIII. 

and  the  three  roots  of  this  cubic  are  the  three  normal  stresses 
of  tension  or  compression,  often  called  the  three  principal 
stresses. 

The  directions  of  these  principal  stresses  with  respect  to  the 
axes  OX,  O  Y,  OZ,  are  given  by  the  values  of  cos  a,  cos  b, 
cos  c,  which  are  found  to  be, 


cos  a  = 
in  which  the  values  of  the  m's  are, 


m9  =  (S,  -  S)(S*  -  S)  -  $*„, 
m,=  (Sx-S)(Sy-  S)  -5^, 
m  =  mt  -[-  mt  -f-  m3 ; 

and  it  will  now  be  shown  that  each  principal  stress  is  perpen- 
dicular to  the  plane  of  the  other  two. 

Let  Slf  S.t ,  5,  be  the  three  roots  of  the  cubic  equation  (30). 
Let  #, ,  £, ,  £,  be  the  angles  which  St  makes  with  the  three  co- 
ordinate axes  OX,  O  Y,  OZ,  and  let  #a ,  £a ,  c,  be  the  angles 
which  5a  makes  with  the  same  axes.  The  angle  between  the 
directions  of  Sl  and  52  is  then  given  by 

cos  0  =  cos  ^  cos  at  +  cos  ^i  cos  ^»  ~t~  cos  c\  cos  c*  • 
Now  in  the  first  set  of  formulas  of  this  article  let  5  be  made 
5,  and  a,  b,  c  be  changed  to  al9  £,,  £,;  let  the  first  equation  be 
multiplied  by  cos  a^ ,  the  second  by  cos  &, ,  and  the  third  by 
cos  r, ;  and  let  the  three  equations  be  added  ;  then 

5, (cos  a,  cos  <za  +  cos  bl  cos  £a  +  cos  cl  cos  ^3), 

is  one  term  in  this  sum.  Again,  let  T  be  made  T9 ,  and  a,  b,  c, 
be  changed  to  a9 ,  bt ,  £, ;  let  the  equations  be  multiplied  by 
cos  al ,  cos  £, ,  cos  r, ,  respectively,  and  added  ;  then, 

5,(cos  a^  cos  «,  -f-  cos  bl  cos  <^,  -["  cos  ^  cos  rs), 
is  one  term  in  the  sum,  while  all  the  other  terms  are  the  same 


ART.   134.  A   NUMERICAL  CASE.  3<DI 

as  before.  Hence  if  Sl  and  5,  are  unequal,  the  factor  in  the 
parenthesis,  which  is  cos  0,  must  vanish  ;  0  is  therefore  a  right 
angle  or  5,  and  St  are  perpendicular.  In  the  same  manner  it 
may  be  shown  that  5,  is  perpendicular  to  both  5,  and  5,. 

The  three  principal  stresses  are  hence  perpendicular  to  each 
other,  and  as  the  only  diameters  of  the  ellipsoid  which  have 
this  property  are  its  axes,  it  follows  that  the  directions  of  the 
principal  stresses  Sl ,  Sa ,  S,  are  those  of  the  axes  of  the  ellip- 
soid of  stress.  These  principal  stresses  thus  give  the  maxi- 
mum normal  stresses  of  tension  or  compression. 

An  interesting  property  of  the  three  rectangular  stresses  S, , 
Sy,  St,  is  that  their  sum  is  constant, whatever  maybe  the  posi- 
tion of  the  coordinate  axes.  For  the  sum  of  the  three  princi- 
pal stresses  5, ,  St ,  St  is  equal  to  the  coefficient  A  in  the  cubic 
of  (30),  and  hence, 

S,+  S,  +  S,  =  S,  +  5,  +  S,; 

that  is,  the  sum  of  the  normal  unit-stresses  in  any  three  rec- 
tangular directions  is  constant. 

Prob.  183.  When  two  principal  stressses  are  equal,  show  that 
the  value  of  each  is  (AB  —  gC)/(2A*  —  6B),  where  A,  B,  C 
are  the  coefficients  in  (30). 

ART.  134.    A  NUMERICAL  CASE. 

To  apply  the  preceding  principles  to  a  particular  case  a 
crank  pin  similar  to  that  investigated  in  Art.  126  may  betaken. 
The  axis  OX  is  assumed  parallel  to  the  axis  of  the  pin,  O  Y 
parallel  to  the  crank  arm,  and  OZ  perpendicular  to  both.  On 
one  side  of  the  crank  pin  near  its  junction  with  the  arm  there 
were  found  the  following  apparent  stresses:  A  cross  shear 
from  the  pressure  of  the  connecting  rod  giving  S^  =  300 
pounds  per  square  inch,  a  shear  due  to  the  transmitted  torsion 
giving  Sxz  =  900  pounds  per  square  inch,  a  flexural  stress  due 


302  APPARENT  AND  TRUE  STRESSES.  CH.  XIII. 

to  the  connecting  rod  giving  S*  =  -j-  800  pounds  per  square 
inch,  a  flexural  stress  due  to  the  transmitted  torsion  giving 
Sx  =  +  l  600  pounds  per  square  inch,  and  two  compressions 
due  to  shrinkage  giving  Sy  =  —  4000  and  Sz  =  —  2  ooo  pounds 
per  square  inch. 

The  two  shears  having  the  same  direction  add  together,  as 
also  the  two  tensions,  and  the  data  then  are, 

Sjfg  =   I  2OO,      S*  =  -f  2  4OO,      Sy  =   —  4  OCX),       Sz  =  —  2  OOO. 

Inserting  these  in  (30)  it  becomes, 

5 3  +  3  6oo52  —  7  840  oooS  —  27  200  ooo  ooo  =  o, 

and  the  three  roots  of  this  are  the  three  principal  stresses. 
To  solve  this  put  5  =  x  —  i  200,  and  it  reduces  to 

x*  —  1 2  160  OOO.T  —  1 1  096  ooo  ooo  =  o. 

As  this  cubic  has  three  real  roots,  it  is  to  be  solved  by  the  help 
of  a  table  of  cosines;  thus  let 

3?*a  =  12  160000,         2ri  cos  30  =  ii  096000000, 

from  which  r  =  2  013  and  cos  30  =  0.680 1.  Then  from  the 
table  30  =  47°  09'  whence  0=15°  43'.  The  roots  now  are 

x^  =  2r  cos  0  =  +  3  880, 

x^  =  2r  cos  (0+  120°)  =  —  2  890, 
x%  =  2r  cos  (0  +  240°)  =  —     990 ; 

and  finally  the  three  principal  stresses  are, 

S,  =  x^  —  i  200  =  +  2  680, 
5,  =  x^  —  i  200  =  —  4  090, 
5,  =  x^  —  i  200  =  —  2  190, 

of  which  Si  is  the  maximum  tension  and  St  the  maximum 
compression. 

These  are  the  apparent  stresses.     Taking  e  =  J  for  steel, 
the  true  principal  stresses  are  now  found  by  (27)  to  be, 

Tt  =  +4770,         T,  =  -  4250,         T,  =  -  i  720, 


ART.  135.  MAXIMUM   SHEARING  STRESSES.  303 

which  shows  that  the  maximum  true  tensile  stress  is  nearly 
double  the  apparent,  while  the  maximum  true  compressive 
stress  is  6  per  cent  greater  than  the  apparent. 

In  any  case  the  method  of  procedure  is  similar.  The  appar- 
ent stresses  being  first  computed  for  the  point  under  consider- 
ation, their  values  establish  the  cubic  equation  (30)  whose 
solution  gives  the  three  principal  apparent  stresses  S,  ,  5,,  S,  ; 
then  from  (27)  the  true  maximum  stresses  of  tension  or  com- 
pression are  found.  It  frequently  happens  that  one  of  the 
principal  apparent  stresses  is  zero  ;  in  this  event  the  last  term 
of  the  cubic  vanishes  and  the  ellipsoid  becomes  an  ellipse. 
For  instance  let  a  bar  be  under  a  tension  S*  and  a  single  cross- 
shear  S*j,.  Then  from  (30)  the  principal  stresses  are 

s,  =  to  + 


the  first  of  which  agrees  with  the  result  given  in  a  different 
notation  in  equation  (13)  of  Art.  75. 

Prob.  184.  Find  the  maximum  true  unit-stresses  of  tension 
or  compression  for  the  data  of  Prob.  121. 

ART.  135.    MAXIMUM  SHEARING  STRESSES. 

As  there  are  certain  planes  upon  which  the  tensile  and  com- 
pressive unit-stresses  are  a  maximum,  so  there  are  certain 
other  planes  upon  which  the  shearing  unit-stresses  have  their 
maximum  values.  In  order  to  determine  these  it  is  well  to 
take  the  axes  of  the  ellipsoid  as  the  coordinate  axes,  and  upon 
the  planes  normal  to  these  there  are  no  shearing  stresses.  The 
stresses  5,,  S9,  S9  will  give  apparent  shearing  stresses  on 
other  planes,  while  Tlt  Ttt  Tt  will  give  the  true  shearing 
stresses. 

Let  i  2  3  in  Fig.  76  be  any  plane  whose  normal  makes  the 
angles  a,  b,  c  with  the  coordinate  axes.  Let  R  be  the  result- 
ant unit-stress  upon  this  plane,  and  a,  ft,  y  be  the  angles  which 


304 


APPARENT  AND  TRUE  STRESSES. 


CH.  XIII. 


it  makes  with  the  same  axes.     The  angle  between  R  and  the 

normal  is  expressed  by, 

cos  0  =  cos  a  cos  a  -\- 

cos  b  cos  ft  +  cos  c  cos  Y* 

and   the  resultant  shearing  unit- 
stress  on  the  plane  is 

R  sin  0  = 


—  cos8  0. 

If  R  be  apparent  stress,  this  is  the 
FIG.  76.  apparent  shearing  stress ;  if  R  be 

true  stress,  this  is  the  true  shearing  stress. 
The  value  of  R,  as  a  true  stress,  is  given  by 

R9  =  (T,  cos  of  +  (T,  cos  b)'  +  (T,  cos  c)\ 

Now,  since  both  R  cos  a  and  Tv  cos  a  are  components  of  R  in 
the  direction  OX,  they  are  equal,  and  hence 

T. 


cos  a  =  — i  cos  #, 


T 

COS  /?=:-—'  COS 
.A. 


cos  y  =  — *  cos 


Substituting  these  in  the  value  of  cos  0,  the  resulting  true 
shearing  unit-stress  is  expressed  by 

T3  =  (T,  cos  of  +  (T,  cos  £)'  +  (r,  cos  <:)'  - 

(T,  cos8  0  +  r,  cos8  3+7;  cos8  <:)', 

by  the  discussion  of  which  the  values  of  #,  b,  c,  which  render  T 
a  maximum,  are  deduced.  Bearing  in  mind  that  the  sum  of 
the  squares  of  the  three  cosines  is  unity,  the  discussion  gives 


(SO 


and  therefore  there  are  six  planes  of  maximum  shearing  stress, 
each  of  which  is  parallel  to  one  of  the  principal  stresses  and 
bisects  the  angle  between  the  other  two.  On  each  of  these 
planes  the  shearing  unit-stress  is  one  half  the  difference  of  the 
principal  unit-stresses  whose  direction  is  bisected. 


ART.  136.  THE  ELLIPSE  OF  STRESS.  305 

The  same  investigation  applies  equally  well  to  the  apparent 
shearing  unit-stresses,  whose  maximum  values  are 

(3 1)'  5  =  ±  4(5,  -  S,),  S=±  i(5,  -  S,),  S=±  4(5,  -  5,), 
a.nd  whose  directions  are  the  same  as  those  of  the  maximum 
true  shearing  unit-stresses.  The  sign  ±  indicates  that  the 
shears  have  opposite  directions  on  opposite  sides  of  the  plane, 
but  in  numerical  work  it  is  always  convenient  to  take  them  as 
positive,  or  rather,  as  signless  quantities. 

As  an  example,  let  a  bar  be  subject  to  a  tension  of  3  ooo 
pounds  per  square  inch  in  the  direction  of  its  length  and  to  a 
compression  of  6000  pounds  per  square  inch  upon  two  oppo- 
site sides.  Here  5,  =  +  3  OCXD,  5,  =  —  6  ooo,  5,  =  o ;  then  from 
(31)'  the  maximum  apparent  shearing  stresses  are, 

5=4  500,        5=3  ooo,         5  =  i  500. 

But  from  (27),  taking  e  as  J,  the  true  tensile  and  compressive 
unit-stresses  are  7\  =  +  5  ooo,  Tt  =  —  7  ooo,  T%  =  +  i  ooo, 
and  then  from  (31)  the  maximum  true  shearing  stresses  are, 

T  =  6  ooo,        T  —  4  ooo,         T  =  2  ooo, 
which  are  33  per  cent  greater  than  the  apparent  ones. 

Prob.  185.  Compute  the  maximum  apparent  and  true  shear- 
ing unit-stresses  for  the  data  given  in  Prob.  121. 

ART.  136.    THE  ELLIPSE  OF  STRESS. 

The  ellipse  of  stress  is  that  particular  case  where  one  of  the 
principal  stresses  is  zero,  in  which  event  the  last  term  of  (30) 
vanishes.  An  instance  of  this  is  where  5,  =  o,  Sy,=  o,  5^., 
=  o,  which  is  that  of  a  body  subject  to  the  normal  stresses 
Sx,  Sy,  and  to  a  cross-shear  S^.  The  cubic  equation  then 
reduces  to 

5'  -  (S,  +  S,)S+  SxSy  -5^  =  0; 

and  the  two  roots  of  this  are  the  two  principal  apparent  stresses 
whose  directions  correspond  to  the  two  axes  of  the  ellipse. 


306 


APPARENT  AND   TRUE  STRESSES. 


CH.  XIII. 


Let  5,  and  5,  be  these  roots,  and  in  Fig.  77  let  OA  and  OB 
be  laid  off  at  right  angles  to  represent  their  values.     Let  an 


B 


FIG.  77- 

ellipse  be  described  upon  the  axes  AA  and  BB,  and  let  0  be 
the  angle  AON  which  any  line  ON  makes  with  OA.  Upon  a 
plane  normal  to  ON  at  O  the  shearing  unit-stress  is 

OS  =  (5,  —  52)  sin  0  cos  0, 
and  the  normal  unit-stress  of  tension  or  compression  is 

ON  =  S,  cos'  0  +  Sa  sin8  0, 
while  the  resultant  unit-stress  is 


OR  =  VS>9  cos9  0  +  5a2  sin'  0. 

The  diagrams  in  Fig.  77  give  graphic  representations  of  these 
values  as  the  angle  0  varies  from  o  to  360  degrees.  In  the 
first  diagram  S,  and  59  are  both  tension  or  both  compression, 
in  the  second  diagram  one  is  tension  and  the  other  compression. 
The  broken  curve  shows  the  locus  of  the  point  N,  and  the 
dotted  curve  the  locus  of  5.  For  every  value  of  0  the  lines 
OS  and  ON  are  at  right  angles  to  each  other,  and  OR  is  their 
resultant. 

As  a  numerical  illustration  take  the  case  of  a  bolt  subject  to 
a  tension  of  2  ooo  and  also  to  a  cross  shear  of  3  ooo  pounds 


ART.  130.  THE  ELLIPSE  OF  STRESS.  307 

per  square  inch.  Here  Sx  =  +  2  ooo,  Sxy  =  3  ooo,  and  Sy  =  o ; 
the  above  quadratic  equation  then  gives  5,  =  +4  160  and 
S9  =  —  2  160  as  the  two  maximum  unit-stresses  of  tension  and 
compression.  The  direction  made  by  5,  with  the  axis  of  the 
bolt,  as  found  by  the  value  of  cos  a  in  Art.  133,  is  about  54^ 
degrees.  From  (31)'  the  maximum  shear  is  3  160  pounds  per 
square  inch.  These  are  the  apparent  maximum  stresses. 

To  find  the  true  maximum  stresses  formulas  (27)  give,  tak- 
ing £  as  the  factor  of  lateral  contraction,  Tt  =  +  4  880, 
7",  =  —  3  550,  7",  =  —  670  pounds  per  square  inch  as  the  prin- 
cipal tensions  and  compressions;  then  from  (31)  the  greatest 
shearing  stress  is  7^=4220  pounds  per  square  inch.  Here 
the  true  maximum  tension  is  17  per  cent  greater  than  the 
apparent,  the  true  compression  is  64  per  cent  greater,  and  the 
true  shearing  stress  is  33  per  cent  greater.  The  true  stresses 
cannot  be  represented  by  an  ellipse,  but  an  ellipsoid  of  internal 
stress  results  of  which  the  second  diagram  in  Fig.  77  may  be 
regarded  as  a  typical  section. 

Cases  can,  however,  be  imagined  in  which  one  of  the  true 
principal  stresses  is  zero.  If  5, ,  5, ,  St  are  the  apparent 
stresses  in  three  rectangular  directions,  it.  is  seen  from  (27) 
that  when  5,  —  eSl  —  eS,  is  zero,  the  true  stress  Tt  is  also  zero. 
For  instance,  let  a  cube  be  under  compression  by  three  nor- 
mal stresses  of  30,  24,  and  18  pounds  per  square  inch  and  let 
e  =  £;  then  T,  =  16,  Tt  =  8,  and  Tt  =  o.  Here  the  ellipse 
of  true  stress  has  its  correct  application  and  there  are  no  true 
stresses  in  a  plane  normal  to  the  plane  of  Tt  and  7",. 

Prob.  1 86.  A  body  is  subject  to  a  tension  of  4000  and  to  a 
compression  of  6000  pounds  per  square  inch,  these  acting  at 
right  angles  to  each  other.  Construct  the  ellipse  of  apparent 
stresses  and  find  the  positions  of  two  planes  on  which  there  are 
no  tensile  or  compressive  stresses. 


308  APPARENT  AND   TRUE   STRESSES.  CH.  XIII. 

ART.  137.     FORMULAS  FOR  TRUE  COMBINED  STRESSES. 

In  the  preceding  articles  it  has  been  shown  how  the  true 
stresses  for  all  cases  may  be  found.  For  the  most  common 
case,  that  of  shear  combined  with  tension  or  compression,  it 
is  well,  in  conclusion,  to  write  the  formulas  by  which  the  true 
maximum  unit-stresses  may  be  computed,  as  these  will  in 
general  give  higher  values  than  the  expressions  deduced  in 
Arts.  76  and  77. 

Let  Sx  be  the  apparent  tensile  unit-stress  and  Sxy  be  the 
apparent  shearing  unit-stress,  the  first  being  found  from  (4), 
and  the  second  from  (3)  for  the  case  of  a  beam  or  from  (12) 
for  the  case  of  a  shaft.  Then  in  the  quadratic  equation  of 
the  last  article  Sy  is  zero,  and  the  two  roots  are 


which  are  the  apparent  maximum  tensile  and  compressive 
unit-stresses  due  to  the  combination  of  Sx  and  Sxy.  Substi- 
tuting these  in  (27)  there  results 


>*y  > 


>xy  > 


which  are  the  true  maximum  unit-stresses,  the  first  being  ten- 
sile and  the  second  compressive.  These  formulas  apply  also 
when  S*  is  a  compressive  stress,  its  sign  being  then  taken  as 
negative.  The  true  maximum  shearing  unit-stress  is  in  both 
cases  given  by  4(7",  —  7!,),  that  is,  by  the  second  term  of  the 
formulas.  For  metals,  the  factor  of  lateral  contraction  e  lies 
usually  between  $  and  J. 

For  instance,  taking  the  numerical  example  on  page  152, 
let  it  be  required  to  find  the  factor  of  safety  of  a  wrought- 
iron  shaft  3  inches  in  diameter  and  12  feet  between  bearings, 


ART.    137.  TRUE    COMBINED   STRESSES.  309 

which  transmits  40  horse-powers  while  making  120  revolu- 
tions per  minute,  and  upon  which  a  *load  of  800  pounds  is 
brought  by  a  belt  and  pulley  at  the  middle.  Here  the 
flexural  stress  on  the  outer  fiber  is  5  400  pounds  per  square 
inch  and  the  torsional  stress  is  4  ooo  pounds  per  square  inch, 
both  acting  at  the  circumference  of  the  shaft.  Thus  %SX  is 


2  700,  and  V^S2X  +  Sly  ls  49°°;   tnen 

T}  =  f  X  2700  +  f  X  4900  =  +  8  300  pounds  per  square  inch, 

and  accordingly  the  factor  of  safety  is  6.6,  whereas  by  consid- 
ering the  apparent  stresses  only  it  was  found  to  be  7.5. 

Finally,  it  may  be  observed  that  the  formula  deduced  in 
Art.  78  for  the  maximum  horizontal  shearing  unit-stress  in  a 
beam  is  not  changed  by  the  introduction  of  the  idea  of  true 
stresses,  since  there  is  no  longitudinal  unit-stress  at  the 
neutral  axis.  Along  this  axis  there  exists  the  horizontal  shear 
and  at  right  angles  to  it  another  of  equal  intensity,  these  com- 
bining to  cause  tensile  and  compressive  unit-stresses  of  the 
same  intensity  in  directions  bisecting  the  lines  of  shear. 

Prob.  187.  Solve  Problems  129  and  130  by  the  application 
of  the  methods  of  this  Chapter. 


STRESSES  IN  GUNS. 


CH.  X.IV, 


CHAPTER  XIV. 
STRESSES  IN  GUNS. 

ART.  138.    LAMP'S  FORMULA. 

Let  a  thick  hollow  cylinder,  shown  in  longitudinal  and 
cross-section  in  Fig.  78,  be  subject  to  a  pressure  pl  on  each 
square  unit  of  the  inner  surface  and  to  a  pressure  /,  on  each 
square  unit  of  the  outer  surface.  The  inner  pressure  may  be 


f 

? 

FIG.  78. 

produced  by  the  expansion  of  a  gas  and  the  outer  pressure  by 
the  atmosphere  or  by  other  causes.  It  is  required  to  deter- 
mine the  internal  stresses  produced  by  these  pressures  at  any 
point  in  the  cylindrical  annulus. 

The  outer  pressure  on  the  end  of  the  closed  cylinder  is 
nr\P*  and  *ne  inner  pressure  on  the  end  is  nr*pr  If  the  inner  be 
greater  than  the  outer  pressure,  as  is  often  the  case,  the  differ- 
ence  of  these,  or  7t(r^pl  —  r,8/,)  is  the  longitudinal  tension  on 
the  annulus.  For  any  part  of  the  cylinder,  not  very  near  the 
end,  this  must  be  uniformly  distributed  over  the  cross-section. 
The  longitudinal  unit-stress  on  the  annulus  is  hence  a  constant 
for  all  points,  and  its  value  is  found  by  dividing  the  total 
stress  by  the  area  of  the  cross-section,  or 


ART.  138.  LAMP'S  FORMULA.  311 

which  may  be  either  tension  or  compression  according  as  the 
numerator  is  positive  or  negative. 

This  longitudinal  stress,  together  with  the  radial  pressures, 
causes  a  longitudinal  elongation  of  the  cylinder,  which  also  is 
to  be  regarded  uniform  for  all  parts  of  the  annulus. 

Let  x  be  the  distance  from  the  center  to  any  point  within 
the  annulus.  Any  elementary  particle  is 
here  held  in  equilibrium  by  the  longi- 
tudinal unit-stress  q,  a  tangential  unit- 
stress  S,  and  a  radial  unit-stress  /.  The 
value  of  /  is  evidently  intermediate  be- 
tween /,  and  /,  ;  in  Fig.  79  it  is,  like  S, 
drawn  as  if  a  tensile  stress.  Now  from  FlG-  7^ 

Art.  129  the  effective  longitudinal  unit-elongation  of  the  cylin- 
der due  to  these  three  stresses  is, 


in  which  e  is  the  factor  of  lateral  contraction  whose  mean 
value  is  about  -J.  But,  as  above  noted,  both  q  and  /  are  con- 
stant for  all  parts  of  the  annulus,  and  it  hence  follows  that 
S-{-p  is  also  a  constant,  or 


which  is  one  equation  between  5  and  p. 

Let  an  elementary  annulus  of  thickness  dx  be  drawn  ;  its 
inner  radius  is  x  and  its  outer  radius  is  x  -\-  dx.  The  pressure 
for  one  unit  of  length  in  a  direction  perpendicular  to  any  diam- 
eter is  px  upon  the  inner  surface  and  (/  +  dp)(x  -f-  dx)  upon 
the  outer  surface  of  this  elementary  annulus.  Thus,  exactly 
as  in  the  case  of  a  thin  pipe  (Art.  9),  the  equation  of  equilib- 
rium is, 

+  dx)  -px  =  Sdx, 


312  STRESSES  IN  GUNS.  CH.  XIV. 

and,  neglecting  the  product  dp  .  dx,  this  reduces  to 

xdp  -\-pdx  —  Sdx, 
which  is  a  second  equation  between  6"  and  /. 

The  solution  of  these  two  equations  gives  for  5  and  /  the 
values 


where  £7,  is  a  constant  of  integration.  To  find  the  values  of 
£7,  and  C9  it  may  be  noted  that  /  becomes  —  pl  when  x  =  rt  , 
and  that  /  becomes  —  •  p^  when  x  =  ra  :  thus, 


and  these  being  inserted  give 


(32) 


.  _  A  ~  A 

P' 


which  are  LAME'S  formulas  for  the  stresses  in  hollow  cylinders 
under  inner  and  outer  pressures.  In  deriving  these  both 
S  and  /  have  been  supposed  to  be  tension.  This  will  be  the 
case  if  the  formulas  give  positive  values  ;  if,  however,  either  3 
or  /  has  a  negative  value  the  stress  is  compression  instead  of 
tension. 

The  tangential  unit-stress  5  is  usually  greater  than  the  radial 
stress  /,  and  is  the  controlling  factor  in  the  design  of  guns.  It 
is  seen  to  increase  as  x  decreases,  and  hence  it  is  the  greatest 
at  the  inner  surface  of  the  cylinder.  It  may  be  either  tension 
or  compression,  depending  upon  the  relative  values  of  the 
radii  and  pressures.  The  radial  pressure  /  is  always  compres- 
sion, its  value  ranging  from  /,  at  the  inner  to  /,  at  the  outer 
surface. 


ART.  139.  A  SOLID  GUN.  313 

As  a  numerical  example  let  a  cylinder  be  one  foot  in  inner 
and  two  feet  in  outer  diameter,  the  inner  pressure  being  600 
and  the  outer  15  pounds  per  square  inch.  Here  rl  =  6  inches, 
r,  —  12  inches,/,  =  600,  /,  =  15,  and  the  formulas  become, 

28  080  28  080 


For  the  inner  surface  x  =  6  inches,  whence  5  =  -f-  960  and 
p  =  —  600  pounds  per  square  inch  ;  at  the  outer  surface  x  ==  12 
inches,  whence  5  =  +  375  and/  =  —  15  pounds  per  square 
inch,  -f-  denoting  tension  and  —  denoting  compression. 

Prob.  1  88.  A  solid  cylinder  is  subject  to  a  unfform  outer 
pressure  of  p%  pounds  per  square  inch.  Prove  that  the  radial 
compression  is  uniform  throughout.  Prove  that  the  tangen- 
tial stress  5  is  compressive  and  equal  to  the  radial  compression 
at  all  parts  of  the  cylinder. 


ART.  139.    A  SOLID  GUN. 

A  gun  tube  without  hoops  is  a  solid  hollow  cylinder,  sub- 
ject upon  the  outer  surface  to  atmospheric  pressure  and  upon 
the  inner  surface  to  the  pressure  of  the  gas  arising  from  the 
explosion  of  the  powder.  The  outer  pressure  /,  is  so  small 
compared  to  the  inner  pressure  /,  that  it  may  be  neglected. 
Then  making  x  equal  to  rl  in  (32)  they  become, 


which  give  the  greatest  tensile  and  compressive  unit-stresses 
caused  by  the  interior  pressure.  The  tangential  tension  and 
the  radial  compression  decrease  as  x  increases,  and  at  the  outer 
surface  where  x  =  rt  their  values  are 

2r/ 

£=    .     '    «A»       /  =  <>> 


STRESSES  IN  GUNS. 


CH.  XIV. 


which  are  the  least  tensile  and  compressive  unit-stresses  caused 
by  the  inner  pressure. 

As  a  special  case  let  the  outer  radius  be  twice  the  inner 
radius  or  r9  =  2r1 .  Then  the  tension  for  the  inner  surface 
becomes  Sl  =  ^pl ,  and  for  the  outer  surfaces  it  is  5t  =  |/A . 


FIG.  80. 

Thus  the  different  parts  of  the  annulus  are  quite  unequally 
stressed,  and  hence  the  material  is  not  utilized  to  the  best 
advantage.  The  shaded  area  in  Fig.  80  shows  the  manner  in 
which  the  tangential  tension  varies  in  this  case  throughout 
the  annulus. 

The  maximum  stress  in  a  solid  gun  is  hence  the  tangential 
tension  at  the  inner  surface  which  is  given  by  the  formula, 


and  this  is  the  expression  frequently  used  in  cases  of  investi- 
gation and  design. 

As  an  example  of  investigation  let  a  cast-steel  gun  have  an 
inner  diameter  of  7.5  inches  at  the  powder  chamber  and  the 
thickness  of  the  tube  be  1.75  inches.  What  is  the  greatest 
tension  produced  when  the  inner  pressure  arising  from  the 
explosion  is  10000  pounds  per  square  inch  ?  Here  r,  =  3.75 
inches,  ra  =  5.50  inches,  /,  =  10000  pounds  per  square  inch. 
Then  from  the  formula  St  is  found  to  be  27  300  pounds  per 


ART.  140.  A  COMPOUND   CYLINDER.  315 

square  inch,  which  is  less  than  the  elastic  limit  of  steel,  and 
hence  not  too  large. 

As  an  example  of  design  let  the  inner  diameter  be  3.25 
inches,  the  inner  pressure  caused  by  the  explosion  15000 
pounds  per  square  inch,  and  the  allowable  working  strengthen 
tension  be  30000  pounds  per  square  inch.  What  should  be 
the  outer  diameter?  Here  rl  =  1.625  inches,  /,  =  15  ooo  and 
Sl  =  30000  pounds  per  square  inch.  Then  solving  the  last 
formula  for  rt  there  results 


from  which  the  outer  radius  rt  is  found  to  be  2.815  inches; 
thus  the  thickness  of  the  tube  is  1.19  inches,  and  its  outer 
diameter  is  5.63  inches. 

Prob.  189.  A  solid  gun  tube  is  6  inches  in  diameter  and  3 
inches  thick.  What  is  the  inner  pressure  that  will  produce  a 
maximum  tangential  tension  of  30000  pounds  per  square 
inch? 

ART.  140.    A  COMPOUND  CYLINDER. 

In  a  solid  gun  the  maximum  tension  occurs  at  the  inner  sur- 
face during  the  explosion,  rising  suddenly  from  o  up  to  its 
greatest  value  5",.  If  now  the  metal  near  the  bore  can  be 
brought  into  compression,  this  must  be  overcome  before  the 
tension  can  take  effect,  and  thus  the  capacity  to  resist  the 
inner  pressure  is  increased.  One  method  of  producing  this 
compression  is  by  means  of  a  hoop,  or  jacket,  shrunk  upon  a 
tube  so  as  to  produce  an  outer  pressure  /,  over  the  surface  of 
the  tube.  This  arrangement  may  be  called  a  hollow  com- 
pound cylinder. 

In  its  normal  state  of  rest  the  inner  cylinder,  or  tube,  has  no 
pressure  on  its  inner  surface  and/,  on  its  outer  surface.  Mak- 


3l6  STRESSES   IN   GUNS.  CH.  XIV. 

mg  A  =  o  in  (32),  and  also  x  =  rl  and  x  =  r,  in  succession, 
there  are  found 


which  are  the  tangential  unit-stresses  at  the  inner  and  outer 
surfaces  of  the  tube  due  to  the  external  pressure/,,.  Both  of 
these  are  compression,  but  the  former  is  numerically  greater 
than  the  latter,  since  2r,2  is  greater  than  ra2  -f-  r*. 

If  the  hoop  is  shrunk  on  so  as  to  produce  a  compressive 
uniSstress  Sc  at  the  inner  surface  of  the  tube,  the  pressure  p% 
upon  the  outer  surface  must  be, 


A=  -57-*' 

and,  if  Sc  be  assumed,  the  shrinkage  may  be  so  regulated  as 
to  produce  this  pressure  /a  in  the  normal  state  of  rest.  Then 
the  tangential  stresses  throughout  the  tube  are  all  compression, 
while  the  radial  pressures  range  from  /,  on  the  outer  surface 
to  o  on  the  inner  surface. 

As  an  example,  let  r,  =  2  inches,  ra  =  3  inches,  and  let  it  be 
required  to  find  the  outer  pressure  which  will  cause  a  tangen- 
tial compressive  unit-stress  of  18000  pounds  per  square  inch 
at  the  inner  surface  of  the  tube.  Here, 


and  hence  the  hoop  must  be  shrunk  on  so  as  to  produce  this 
outer  pressure  on  the  hoop. 

When  the  gun  is  fired  the  explosion  of  the  powder  causes 
an  internal  tangential  tension  5  given  by  (32),  whose  greatest 
value  is  at  the  inner  surface  of  the  tube.  Making  x  =  rl  ,  this 
tensile  unit-stress  is, 

(32y  St=BfcL 


ART.   141.  CLAVARINO'S   FORMULAS.  317 

which  is  LAME'S  formula  for  the  investigation  of  the  tube  of  a 
compound  gun.     This  tension  first  overcomes  the  initial  com 
pression  Sc  due  to  shrinkage,  so  that  the  effective  tension  at 
the  bore  during  firing  is  5,  —  Sc. 

For  example,  let  a  tube  whose  inner  and  outer  diameters  are 
4  inches  and  6  inches  be  hooped  so  that  a  tangential  com- 
pression of  18000  pounds  per  square  inch  is  caused  at  the 
bore,  while  the  inner  pressure  caused  by  the  explosion  is  25  ooo 
pounds  per  square  inch.  It  is  required  to  find  the  tangential 
stress  at  the  bore  during  the  explosion.  Here  rl  =  2  and 
r,  =  3  inches,/,  =  25  ooo,  and/,  =  5  ooo,  as  seen  above.  Then 
Sl  =  47  ooo  pounds  per  square  inch  is  the  tension  due  to  the 
explosion,  but  before  this  can  take  effect  the  initial  compres- 
sion of  18000  pounds  must  be  overcome.  Hence  the  result- 
ant tension  at  the  bore  during  the  firing  is  47  ooo  —  18  ooo  = 
29  ooo  pounds  per  square  inch. 

If  this  tube  be  without  a  hoop  the  tension  at  the  inner  sur 
face,  found  by  the  method  of  the  last  article,  is  57  200  pounds 
per  square  inch.  The  very  great  advantage  of  the  hoop  in 
diminishing  the  internal  stresses  during  the  firing  is  hence 
apparent. 

Prob.  190.  A  gun-tube  3  inches  in  diameter  and  1.5  inches 
thick  is  hooped  so  that  the  tangential  compression  on  the 
inner  surface  is  30  ooo  pounds  per  square  inch.  What  inner 
pressure/,  will  produce  a  resultant  tangential  tension  on  the 
inner  surface  of  30  ooo  pounds  per  square  inch  ? 

ART.  141.    CLAVARINO'S  FORMULAS. 

The  preceding  method  of  investigating  the  strength  of  gun 
tubes  is  defective  in  that  the  two  stresses  5  and  /  of  formula 
(32)  are  only  apparently  the  internal  stresses.  It  was  shown 
in  Art.  71  and  also  in  Art.  129  that  the  true  internal  stresses 
are  those  corresponding  to  the  deformations  produced.  These 


318  STRESSES  IN  GUNS.  CH.  XIV. 

deformations  are  influenced  by  the  factor  of  lateral  contrac- 
tion of  the  material,  which  for  steel  and  gun  metal  is  usually 
taken  as  J  (Art.  128). 

At  any  point  in  the  annulus  (Fig.  79)  the  apparent  tangen- 
tial, radial,  and  longitudinal  unit-stress  are  S,  p,  and  £,  respec- 
tively. Let  T  be  the  true  tangential  unit-stress  ;  then  from  (27) 
of  Art.  129  its  value  is, 

r=s-i/-fe. 

Inserting  in  this  the  values  of  S,/,  and  q  from  Art.  138,  it  re- 
duces to, 

t,,\        T-  r-&-ll&  +  -JfiSL  A  -A 

•3(n'-O     3(r,'  -*-,')•    *• 

which  is  CLAVARINO'S  principal  formula  for  the  investigation 
and  design  of  guns. 

This  formula  shows,  as  before,  that  the  tangential  stress  is 
greatest  at  the  inner  surface  of  the  cylinder.  Making  x  =  r^  , 
this  maximum  tension  is 


T    - 


which  is  the  practical  formula  for  the  discussion  of  the  most 
common  cases.  7i  may  be  either  tension  or  compression,  de- 
pending upon  the  relative  values  of  the  pressures  and  radii. 

CLAVARINO'S  formulas  are  now  frequently  used  in  the 
investigation  and  design  of  guns,  instead  of  those  of  LAME. 
In  order  to  compare  them,  the  particular  case  where  the  outer 
diameter  is  double  the  inner  diameter  may  be  considered. 
Here  ra  =  2rt  ,  and  (32)'  and  (33)'  reduce  to 

c  __  5A  —  8/«  T  -  *#'  -  2QA 

Of    ----  2     —  --  . 

3  9 

Now  if  /3  =  o,  the  first  formula  gives  a  smaller  stress  than  the 
second  ;  if  /,  =  plt  the  first  gives  a  stress  three  times  as  large 
as  the  second  ;  if  /,  =  o,  the  first  gives  a  little  larger  stress 


ART.  142.  BIRNIE'S  FORMULAS.  319 

than  the  second.  Thus,  since  the  second  formula  gives  un- 
doubtedly a  better  representation  of  the  true  stress  than  the 
first,  it  follows  that  LAMP'S  method  errs  on  the  side  of  danger 
in  a  solid  gun  and  on  the  side  of  safety  in  a  hooped  tube. 

The  value  £  here  used  as  the  coefficient  of  lateral  contrac- 
tion is  that  employed  in  the  United  States  by  both  the  Army 
and  the  Navy  in  ordnance  formulas,  and  also  generally  in 
Europe.  In  France,  however,  the  value  e  =  J  is  adopted. 

Prob.  191.  Solve  Problem  189  by  the  formulas  of  this  ar- 
ticle, and  compare  the  two  results. 

ART.  142.    BIRNIE'S  FORMULAS. 

The  preceding  articles  present  an  outline  of  the  methods  of 
investigating  stresses  in  guns  by  the  formulas  of  LAM£  and 
CLAVARINO.  The  formulas  of  LAM£  refer  to  apparent  stresses 
only ;  those  of  CLAVARINO,  although  referring  to  true  stresses, 
contain  an  error  which  renders  them  not  strictly  correct  for 
hooped  guns.  This  error  lies  in  taking  for  the  longitudinal 
unit-stress  q  the  value  given  in  first  equation  of  Art.  138.  That 
value  of  q  is  correct  for  a  hollow  cylinder  subject  to  pressure 
upon  its  ends  as  well  as  upon  its  curved  surfaces.  For  a  gun, 
however,  q  has  a  different  value.  When  the  gun  is  at  rest  q  is 
zero,  for  then  no  exterior  longitudinal  forces  act  upon  it. 

For  the  investigation  of  a  gun  at  rest  the  true  tangential 
•  stress  T  should  be  deduced  by  making  q  =  o.     Thus,  in  Art. 
141  the  correct  value  of  T  is  5  —  \p,  or, 


3  W  -  o  r        • 

which  is  BIRNIE'S  formula  for  the  investigation  and  design 
of  hooped  guns.     Making  x  =  rt  ,  this  becomes, 


/7  ,y  T  - 

(34)  r'^~        3(r.'-0 


320  STRESSES  IN  GUNS.  CH.  XIV. 

which  is  the  tangential  unit-stress  at  the  inner  surface  of  a 
hoop  whose  radii  are  rl  and  ra.  These  formulas  are  used  in 
the  Ordnance  Bureau  of  the  United  States  army,  not  only  for 
hoops,  but  for  jackets  and  tubes,  both  at  rest  and  during  the 
firing. 

Strictly  speaking,  BIRNIE'S  formulas  apply  only  to  hoops 
and  tubes  upon  which  the  exterior  longitudinal  stress  q  is  zero. 
For  a  solid  gun,  or  for  a  tube  attached  to  the  breech  block,  a 
more  correct  formula  may  be  found  by  considering  the  actual 
value  of  q  due  to  the  inner  pressure.  Here  the  longitudinal 
pressure  is  nr^pl  ,  and  this  produces  longitudinal  tension  upon 
the  area  n(r*  —  r,a),  so  that 

q  ~  r**—lr* 

*9  T\ 

is  the  apparent  longitudinal  unit-stress.  Then  the  true  tan- 
gential stress  T  is  .S  —  \p  —  \q,  or, 


r?r;        *.  -A 
'"       • 


and  making  in  this  x  =  rl  ,  it  becomes 


which  is  the  true  tangential  unit-stress  upon  the  inner  surface 
of  the  bore. 

To  compare  the  formulas  of  CLAVARINO  and  BIRNIE  the 
particular  case  where  ra  =  2r,  may  be  considered.  Then  (33)' 
and  (34)'  reduce  to 


__  i/A  ~  so/,         T  _  i8A  -  24A 
"T"  "T" 

Now,  if  A  =  o>  as  f°r  a  solid  gun  during  firing,  the  second  for- 
mula gives  a  stress  6  per  cent  larger  than  the  first  ;  if  A  =  °> 
as  for  a  hooped  gun  at  rest,  the  second  gives  a  stress  25 


ART.  143.  HOOP  SHRINKAGE.  321 

per  cent  greater  than  the  first.     Thus  for  this  case,  at  least, 
CLAVARINO'S  formulas  give  the  stresses  somewhat  too  low. 

Prob.  192.  Solve  Problem  189  by  the  formulas  of  this  ar- 
ticle and  compare  the  results  with  those  of  Problem  191. 

ART.  143.    HOOP  SHRINKAGE. 

Let  A  be  the  elongation  or  contraction  of  any  radius  x  due  to 
inner  or  outer  pressure,  then  2n\  is  the  elongation  or  contrac- 
tion of  any  circumference  2nx.  Now  2n\/2nx  is  the  change 
in  the  circumference  per  unit  of  length  due  to  the  unit-stress 
T\  hence  \/x  =  T/E,  or 


is  the  change  in  length  of  any  radius  to  the  circle  where  the 
tangential  unit-stress  is  71  If  x  =  rt  the  deformation  A,  is 
that  of  the  radius  of  the  bore  due  to  the  unit-stress  Tv:  if 
x  =  r9  the  change  A,  is  that  of  the  outer  radius  of  the  tube 
where  the  unit-stress  is  Tf 

Suppose  a  compound  cylinder  to  be  formed  by  shrinking  a 
hoop  upon  a  tube.  The  inner  radius  of  the  tube  is  rl  and  its 
outer  radius  r,  ;  the  inner  radius  of  the  hoop  is  r,  and  its  outer 
radius  rt.  In  consequence  of  the  shrinkage  the  radial  pressure 
/>a  is  produced  between  the  two  surfaces  ;  this  causes  the  tube 
to  be  under  tangential  compression  and 
the  hoop  to  be  under  tangential  tension. 
It  is  required  to  find  these  stresses  when 
the  original  inner  radius  of  the  hoop  is  less 
than  that  of  the  outer  radius  of  the  tube 
by  the  amount  A. 

Let  A,  be   the  decrease   in   the   outer 
radius  of  the  tube  and  A,'  the  increase  in 

the  inner  radius  of  the  hoop  ;  then  A  =  A,  -f-  ^/«  ^n  Fig-  8  1, 
which  is  much  exaggerated,  cd  represents  A,  and  be  repre- 


322  STRESSES  IN  GUNS.  CH.  XIV. 

sents  \J.  Also,  let  Tt  be  the  tangential  compression  at  the 
•outer  surface  of  the  tube  due  to  the  shortening  Aa,  and  let 
77  be  the  tangential  tension  at  the  inner  surface  of  the  hoop 
due  to  the  elongation  A/.  Then 


or 

T,  +  T,'  =  Q, 

't 

which  gives  one  equation  between  7!,  and  7",'. 

Formula  (34)  is  applied  to  the  tube  by  making  pl  =  o  and 
x  =  rt  ;  thus  the  tangential  compression  Js, 


Formula  (34)  is  applied  to  the  hoop  by  replacing  /,  by  /,  ,  /, 
by  o,  r,  by  r,  ,  and  rt  by  r,  ;  then  making  x  =  r,  ,  there  results, 

2 
~ 


which  is  the  tangential  tension.     Dividing  the  first  of  these  by 
the  second,  gives 

L.f  =  a-    or    r,£=7>, 

which  is  a  second  equation  between  7",  and  71/. 
The  solution  of  these  equations  gives  the  values 
£A       a  ,  _E\.       b 

'   r,  'a  +  6'  '   r,  ' 

in  which  a  and  b  are  the  known  quantities, 
_2   2rta  +  r,g  ,__^    r,a 

"  * 


and  thus  the  tangential  compression  at  the  outer  surface  of  the 
tube  and  the  tangential  tension  at  the  inner  surface  of  the  hoop 
may  be  computed.  The  tangential  compression  at  the  bore  is, 


ART.  143.  HOOP  SHRINKAGE.  323 

however,  greater  than  7"a,  and  it  may  be  found  from  (34)'  by 
substituting  the  value  of/,,  now  known  ;  thus, 


T 

J 


which  is  the  greatest  compression  in  the  tube. 

As  a  numerical  example  let  a  compound  cylinder  be  formed 
of  a  steel  tube  whose  inner  radius  is  3  inches  and  outer  radius  5 
inches,  with  a  steel  hoop  whose  thickness  is  2  inches.  It  is  re- 
quired to  find  the  stresses  produced  when  the  original  difference 
between  the  outer  radius  of  the  tube  and  the  inner  radius  of 
the  hoop  is  0.004  inches.  First,  the  sum  of  the  two  tangential 
stresses  at  the  surface  of  contact  is, 

E\  _   30  ooo  ooo  X  0.004  _ 

--  --  —  24  ooo. 

r*  5 

Second,  since  rt  =  3,  r,  =  5,  and  rt  =  7  inches, 


—  £  l8  +  25  __  43         h  —  -  25  +  98  _  82 
"  3  25  -  9  ""  24'  "349-25  "24 

Third,  the  tangential  compression  at  the  outer  surface  of  the 
tube  is, 

T%  =  —  X  24  ooo  =  8  260  pounds  per  square  inch. 

43  -r  82 

Fourth,  the  tangential  tension  at  the  inner  surface   of  the 

hoop  is, 

ft? 

TV  =  -  -—  X  24  ooo  =15  740  pounds  per  square  inch. 
43  +  82 

Thus  it  is  seen  that  the  hoop  tension  is  nearly  double  the  com. 
pression  on  the  outer  surface  of  the  tube.  At  the  bore  of  the 
tube,  however,  the  tangential  compression  is  found  to  be  14  400 
pounds  per  square  inch. 

The  decrease  in  the  outer  radius  of  the  tube  is  next  com- 
puted; thus, 

A,  =  ——^  =  0.00138  inches, 


324  STRESSES   IN   GUNS.  CH.  XIV. 

and  the  increase  in  the  inner  radius  of  the  hoop  is, 

T'r 

A,'  =  — ~  —  0.00262  inches. 
h 

Hence  if  the  radius  of  the  common  surface  of  contact  is  to  be 
exactly  5  inches  after  the  shrinkage,  the  tube  should  be  turned 
to  an  outer  radius  of  5.0014  inches,  and  the  hoop  to  an  inner 
radius  of  4.9974  inches.  The  radius  of  the  bore,  however,  will 
then  be  less  than  3  inches  by  the  quantity, 

T 

A,  =  —^  =  0.00144  inches, 
£, 

and  hence  if  its  final  radius  is  to  be  exactly  3  inches,  it  must 
be  turned  to  a  radius  of  3.0014  inches. 

If  this  example  be  solved  by  using  the  formulas  of  CLAVA- 
RINO  instead  of  those  of  BlRNlE,  the  following  values  will  be 
found:  J",  =  7  030,  71/  =  16970,  7^  —  14400  pounds  per 
square  inch;  Aa  —  0.00117,  \'  =  0.00283,  and  ^,  —  0.00144 
inches.  The  shrinkages  thus  agree  within  0.0002,  which  is  as 
close  as  measurements  can  be  relied  upon. 

Prob.  193.  A  solid  steel  shaft,  6  inches  in  radius,  is  to  be 
hooped  so  that  the  greatest  tensile  stress  in  the  hoop  and  the 
greatest  compressive  stress  in  the  shaft  shall  be  15  ooo  pounds 
per  square  inch.  Find  the  thickness  of  the  hoop  and  the 
radius  to  which  it  should  be  turned. 


ART.  144.    HOOPED  GUNS. 

A  hooped  gun  should  be  so  constructed  that  neither  the 
stresses  due  to  hoop  shrinkage  nor  those  developed  during  the 
firing  shall  exceed  the  elastic  limit  of  the  material.  The  sim- 
ple case  of  a  tube  with  one  hoop  can  here  only  be  considered. 
If  the  radii  be  given,  as  also  the  inner  pressure  /,  due  to  the 
explosion,  it  may  be  desired  to  find  the  shrinkages  so  that  this 
requirement  will  be  fulfilled.  As  pl  is  very  large,  it  is  desir- 


ART.  144. 


HOOPED   GUNS. 


325 


able  that  the  given  dimensions  should  be  such  as  to  require  the 
least  amount  of  material. 

The  condition  of  minimum  amount  of  material  will  be  in 
general  fulfilled  when  the  stresses  during  the  explosion  are  as 
great  as  allowable  and  as  nearly  equal  as  possible.  The  dia- 
gram in  Fig.  82  represents  the  distribution  of  the  internal 
stresses  under  this  supposition.  O  is  the  center  of  the  gun, 

b 


FIG.  82. 

OA  the  inner  radius  rl ,  while  AB  is  the  thickness  of  the  tube 
and  BC  that  of  the  hoop.  The  shaded  areas  show  the  stresses 
due  to  the  hoop  shrinkage,  Aa  and  Bb  being  the  tangential 
compressions  Tl  and  7",  of  the  last  article,  while  Bb'  is  the  tan- 
gential tension  71/,  and  Cc  is  the  tangential  tension  at  the  outer 
surface  of  the  hoop.  When  the  explosion  occurs  the  two 
cylinders  are  thrown  into  tangential  tension,  Aal  and  Bbv 
being  those  at  the  Inner  surfaces  of  the  tube  and  hoop.  The 
above  principle  indicates  that  both  Aat  and  Bbl  should  be 
equal  to  the  maximum  allowable  unit-stress  7*«,  which  for 
guns  is  often  taken  nearly  as  high  as  the  elastic  limit  of  the 
material. 

In  designing  a  gun  the  radius  of  the  bore  and  the  thickness 
of  the  tube  may  be  assumed,  and  it  may  be  required  to  find  the 
thickness  and  shrinkage  of  the  hoop  so  that  the  stresses  Aay 
Aal ,  and  Bbl  in  Fig.  82  are  each  equal  to  the  elastic  limit  of 
the  material.  Or,  given  the  radius  of  the  bore  and  the  outer 


326  STRESSES  IN  GUNS.  CH.  XIV. 

radius  of  the  hoop,  it  may  be  required  to  find  the  intermediate 
radius  under  the  same  conditions.  These  problems  can  be 
solved  as  well  as  more  complex  ones  relating  to  guns  with 
several  hoops. 

The  stresses  in  guns  are  greatest  near  the  powder  chamber, 
since  the  gas  expands  and  its  pressure  decreases  as  the  pro- 
jectile moves  outward.  Hence  the  hoops  increase  in  number 
toward  the  powder  chamber,  each  being  shorter  than  the  one 
beneath  it.  Guns  with  seven  or  more  hoops  have  been  built. 
Wire  has  been  used  instead  of  hoops,  but  not  with  good  results. 

The  longitudinal  stress  is  mostly  borne  by  the  tube  if  that 
be  attached  to  the  breech  block.  In  this  case  the  longitudinal 
internal  pressure  during  firing  is  7tr*p^  and  this  divided  by 
n(r*  —  r?)  gives  the  apparent  longitudinal  unit-stress  q.  This, 
however,  is  decreased  by  the  influence  of  the  tangential  and 
radial  pressures.  Thus,  from  Arts.  129  and  138, 


which  is  the  true  longitudinal  unit-stress  on  the  tube.  When 
the  jacket  is  attached  to  the  breech  block,  which  is  more  often 
the  case,  it  is  subject  to  the  inner  pressure  /„  from  the  tube, 
to  the  outer  pressure  /,  from  the  hoop,  and  it  carries  the 
entire  longitudinal  stress  7tr?pl  ;  thus 


T 


is  the  longitudinal  unit-stress  on  the  jacket. 

Reference  is  made  to  the  excellent  work  of  MEIGS  and 
INGERSOLL,  The  Elastic  Strength  of  Guns  (Baltimore,  1885), 
for  a  detailed  presentation  of  the  formulas  and  methods  used 
in  the  United  States  Navy  for  the  design  of  guns.  The  for- 
mulas used  by  the  Ordnance  Bureau  of  the  Army  will  be 


ART.  144.  HOOPED  GUNS.  327 

found  set  forth  in  a  thorough  manner  in  STORY'S  Elements 
of  the  Elastic  Strength  of  Guns  (Fort  Monroe,  1894). 

Prob.  194.  Prove  that  a  gun  tube  with  one  hoop  is  most 
advantageously  designed  when  the  common  radius  of  tube  and 
hoop  is  a  mean  proportional  between  the  other  two  radii.  (To 
solve  this  problem  derive  an  expression  for  /,  in  terms  of  rlf 
r9,rt,  and  Te.  Then  the  most  advantageous  value  of  r,  is 
that  which  renders/,  a  maximum.) 


328  PLATES,   SPHERES,   AND   COLUMNS.  ClI.  XV. 


CHAPTER   XV. 
PLATES,  SPHERES,  AND   COLUMNS. 

ART.  145.    CIRCULAR  PLATES. 

Let  a  circular  plate  of  radius  r  and  uniform  thickness  d  be 
subject  on  one  side  to  a  pressure  /  on  each  square  unit  of 
area  and  be  supported  or  fixed  around  the  circumference.  The 
head  of  a  cylinder  under  the  pressure  of  water  or  steam  is  a 
circular  plate  in  such  a  condition.  The  maximum  stress 
caused  by  the  flexure  will  evidently  be  at  the  middle,  and  this 
is  required  to  be  determined. 

As  the  simplest    case    let  the  plate  be   merely  supported 
around  the  circumference.     The  total  load  on  the  plate  being 
7tr*p,  the   total  reaction   of    the    support 
is  also    nr*p,  or  the  reaction  per  linear 
unit  is  \rp.     Now  let  a  strip  having  the 
small  width  b  be  imagined  to  be  cut  out 
of  the  plate,  so  that  its  central  line  co- 
incides  with   a   diameter.     The   reaction 
UUIUUUU         at  each  end  of  this  strip  is  b.^rp,  and  the 


j^mT**:nj jt rp?       load  on  the  strip  is  b.2r.p.     The  sum  of 

the  two  reactions  being  only  one  half  the 

load,  an  upward  shearing  force  equal  to 

b.r.p  must  act  along  the  sides  of   the  strip  to  maintain  the 

equilibrium.     At   the  center   of   the   circle   there   can  be  no 

shearing  stress  and  the  most  probable  assumption  regarding  its 

distribution  on  the  sides  of  the  strip  is  to  take  it  as  varying 

uniformly  from  the  center  to  the  circumference,  as  shown  by 

the  dotted  lines  in  Fig.  83. 


ART.  145.  CIRCULAR  PLATES.  329 

The  strip  whose  width  is  b  and  length  2r  is  thus  a  beam 
acted  upon  by  two  vertical  reactions,  each  equal  to  tyrp,  a 
downward  load  2brpy  and  two  vertical  shears  on  the  sides 
each  equal  to  \brp.  The  bending  moment  at  the  middle  of 
this  imaginary  beam  hence  is, 

M—  \brp  .  r  +  \brp  .  \r  —  brp  .  \r  =  \br*p, 

and  the  maximum  unit-stress  on  the  upper  or  lower  fiber  at 
the  middle  of  the  strip  is, 

_  Me  _  6M  _       r^ 
S>~    f   =  bd'-2p<T 

This  value  of  Sl  is  not  the  real  horizontal  fiber  stress  at  the 
center  of  the  circle,  but  only  the  apparent  stress  due  to  con- 
sidering the  elementary  strip.  At  the  center  the  fiber  stresses 
are  acting  in  all  directions.  If  a  second  strip  be  passed  in 
Fig.  83  at  right  angles  to  the  first,  a  stress  5,  equal  in  value 
but  normal  in  direction  to  the  first  will  be  found.  The  true 
fiber-stress  T  will  be  determined  from  the  principle  of  Art. 
129,  taking  into  account  the  factor  of  lateral  contraction  e; 
thus  on  the  upper  side  of  the  plate, 

T=(S,-eS,+  ef) 

and  on  the  lower  side  of  the  plate, 


which  is  the  value  to  be  used,  since  rupture  will  generally  be- 
gin  on  the  tensile  side. 

For  iron  and  steel  the  mean  value  of  the  factor  of  lateral 
contraction  e  is    .     Hence 


is  the  practical  formula  for  the  discussion  of  iron  and  steel 
plates  under  a  uniform  pressure  when  simply  supported  at  the 
circumference.  The  unit-pressure  that  such  a  plate  can  carry 


330  PLATES,   SPHERES,  AND  COLUMNS.  CH.  XV. 

with  a  given  unit-stress  hence  varies  directly  as  the  square  of 
its  thickness  and  inversely  as  the  square  of  its  radius. 

The  more  common  case  of  a  circular  plate  fixed  around  its 
circumference  cannot  be  solved  without  determining  the  elastic 
curve  into  which  a  diameter  deflects.  The  investigation  is  too 
difficult  and  lengthy  for  this  elementary  book,  but  it  can  be 
said  that  the  general  conclusion  is  that  the  true  effective  unit- 
stress  T  is  about  three-fourths  of  that  for  the  supported  plate. 
Using  this  result  the  coefficient  in  the  last  formula  will  be 
unity,  or 


is  the  practical  formula  for  iron  and  steel  plates  under  uniform 
pressure  when  fixed  around  the  circumference. 

Assuming  a  safe  working  stress  T,  the  proper  thicknesses 
of  plates  under  uniform  pressure  then  are 


=r~        and  d= 


the  first  being  for  supported  and  the  second  for  fixed  circum- 
ference. For  example,  let  a  fixed  cast-iron  cylinder  head  of  3 
feet  diameter  be  required  to  sustain  a  uniform  pressure  of  250 
pounds  per  square  inch,  so  that  the  maximum  tensile  unit- 
stress  T  may  be  3  600  pounds  per  square  inch  ;  then 


which  is  the  required  thickness. 

The  formulas  derived  by  GRASHOF  from  a  more  extended 
theoretical  analysis,  are, 

2p 


and        d= 

for  the  thickness  of  supported  and  fixed  plates  respectively. 


ART.    146.  ELLIPTICAL  PLATES.  331 

The  second  of  these,  applied  to  the  above  numerical  example, 
gives  d  =  3$  inches. 

Prob.  195.  Deduce  a  formula  for  a  circular  plate  sup- 
ported around  the  circumference  and  carrying  a  single  load  P 
at  the  center. 

ART.  146.    ELLIPTICAL  PLATES. 

Elliptical  plates  are  commonly  used  for  the  covers  of  man- 
holes in  boilers  and  stand-pipes.  Let/  be  the  uniform  unit- 
pressure  on  the  plate,  a  the  major  axis  and  b  the  minor  axis  of 
the  ellipse.  It  is  required  to  find  the 
maximum  unit-stress  5  on  the  tensile 
side  of  the  plate. 

Taking  the  case  where  the  plate  is 
simply  supported  around  the  circumfer- 
ence, let  two  elementary  strips  be 

drawn  as  in  Fig.  84,  one  along  the  major  axis  and  the  other 
along  the  minor  axis.  Let  Wl  and  Wt  be  the  loads  on  these 
strips,  and  Rt  and  Rt  the  reactions  at  their  ends.  At  the  center 
they  have  a  common  deflection,  which  by  Art.  36  is 


__  _  W? 

~'mEl  ~ 


and  hence  Wj?  =  WJ?\  or,  since  the   reactions  are  proper- 
tional  to  the  loads, 


j   =  or 

that  is,  the  reactions  at  the  ends  of  the  axes  are  inversely  as 
the  cubes  of  the  lengths  of  the  axes.  Hence  the  total  weight 
±nabp  is  not  uniformly  distributed  on  the  support  around  the 
circumference,  but  the  greatest  reaction  per  linear  unit  will  be 
found  at  the  ends  of  the  minor  axis  and  the  least  at  the  ends 
of  the  major  axis.  It  should  hence  be  expected  that  the  fiber 
stresses  near  the  center  are  the  greatest  in  directions  parallel 


332  PLATES,   SPHERES,  AND  COLUMNS.  ClI.  XV. 

to  the  minor  axis,  and  that  in  case  of  rupture  a  crack  would 
begin  at  the  center  and  run  along  the  major  axis  ;  this  is  veri- 
fied by  tests. 

A  satisfactory  solution  of  this  very  difficult  problem  has 
not  been  made.  From  the  discussion  given  by  BACH  (Elas- 
ticitat  und  Festigkeit,  1894)  the  following  approximate  formula 
results,  which  may  advantageously  be  used  for  elliptical  plates 
in  the  absence  of  other  rules  :  t 

Ca*b*p 
"  (a*  -f  tf}d" 

in  which  5  is  th~  maximum  unit-stress  at  the  center,  d  is  the 
thickness  of  the  plate,  and  C  is  a  number  which  is  probably 
about  f  when  the  circumference  is  supported  and  about  J 
when  it  is  fixed. 

Prob.  196.  A  common  proportion  for  manhole  covers  is  to 
make  a  —  i.$b.  If  the  thickness  be  f  inches,  the  major  axis 
16  inches,  and  the  material  wrought  iron,  find  the  safe  pressure 
per  square  inch. 

ART.  147.    RECTANGULAR  PLATES. 

A  rectangular  plate  of  length  a  and  breadth  £,  subject  to  a 
uniform  pressure^  per  square  unit,  distributes  that  pressure 
over  the  supports  in  a  similar  manner  to  the  elliptical  plate. 
The  reaction  per  linear  unit  is  less  on  the  ends  than  on  the 
sides,  and  is  greater  at  the  middle  of  the  ends  and  sides  than 
near  the  corners.  Rupture  tends  to  occur  near  the  center  and 
parallel  to  the  longer  side.  The  approximate  formula  derived 
by  BACH  for  such  plates  is, 

e_        C 
~ 


in  which  5"  is  the  maximum  unit-stress  at  the  middle,  and  d  is 
the  thickness  of  the  plate.  The  value  of  C  as  found  by  BACH, 
by  experiments  on  square  plates,  ranged  from  f-  to  f  ,  according 


ART.  148.  HOLLOW  SPHERES.  333 

as  the  condition  of  the  edges  approached  that  of  a  mere  sup- 
port or  a  state  of  fixedness. 

For  a  square  plate  whose  side  is  a  this  formula  gives 


c  _  and          c  _ 

"  - 


for  free  and  fixed  supports  respectively.  A  theoretic  analysis 
by  GRASHOF  for  square  plates  fixed  at  the  middle  of  each  side 
gives  the  true  unit-stress  as 


<*>-  gd> 

if  the  factor  of  lateral  contraction  e  be  taken  at  -J,  as  usual  for 
iron  plates  (Art.  128). 

While  the  numerical  coefficients,  as  deduced  by  different 
authors,  vary  somewhat,  it  is  well  established  that  the  unit- 
stress  at  the  middle  of  the  plate  varies  directly  as  its  area  and 
the  unit-pressure  /,  and  inversely  as  the  thickness  of  the  plate. 
The  strength  of  a  plate  as  measured  by  the  pressure  /  that  it 
can  carry  varies  directly  as  the  square  of  the  thickness  and  in- 
versely as  the  area. 

Prob.  197.  Prove  that  the  maximum  unit-stress  caused  by  a 
given  uniform  load  Wis  independent  of  the  size  of  the  plate. 

ART.  148.    HOLLOW  SPHERES. 

Hollow  spheres  are  used  in  certain  forms  of  boilers  under 
inner  steam-pressure.  The  ends  of  steam  and  water  cylinders 
are  sometimes  made  hemispherical  instead  of  plane,  in  order 
to  avoid  flexure  (Art.  145).  If  the  thickness  of  the  sphere  is 
small  compared  to  its  radius,  the  investigation  is  simple.  Let 
r  be  the  radius  and  t  the  thickness.  Let  p  be  the  inner  press- 
ure per  square  unit,  and  5  the  tensile  unit-stress  on  the  an- 
nulus.  Then  on  any  great  circle  the  total  pressure  is  nr'p,  and 


334  PLATES,    SPHERES,   AND   COLUMNS.  CH.  XV, 

this  is  resisted  by  the  tension  2nrtS  in  the  section  of  the  an 
nulus.     Equating  these  gives 

rp  =  2/5,        or         S  =  2, 

which  is  the  formula  generally  used  for  thin  spheres  under 
inner  pressure.  But  in  strictness  5  is  the  apparent  stress, 
while  another  equal  in  intensity  acts  at  right  angles  to  it. 
Thus  from  Art.  129  the  true  stress  on  the  inner  surface  is 


which  gives  the  maximum  true  unit-stress,  since  that  on  the 
outer  surface  is  f  S.  The  usual  formula  thus  errs  on  the  side 
of  safety. 

The  investigation  of  a  thick  hollow  sphere  under  inner  and 
outer  pressure  will  be  similar  to  that  of  the  thick  cylinder  in 
Art.  138.  Let  rl  be  the  inner  and  rt  the  outer  radius,  pl  and 
/,  being  the  corresponding  pressures  per  square  unit  of  surface. 
Fig.  79  of  Art.  138  may  represent  a  partial  section  of  the 
sphere,  x  being  the  radius  of  any  elementary  annulus  where 
the  radial  unit-stress  is  /  and  the  tangential  unit-stress  is  5. 
From  the  symmetry  of  the  sphere  it  is  seen  that  another  stress 
5  acts  at  right  angles  to  the  one  shown  in  the  figure.  Thus 
an  elementary  particle  at  any  position  in  the  annulus  is  held  in 
equilibrium  by  three  principal  stresses  /,  S,  and  S.  The  sum 
of  these  is  regarded  as  constant  throughout  the  annulus,  or 


is  one  equation  between  5"  and  /. 

Now  the  inner  pressure  on  a  great  circle  whose  radius  is  x 
is  nx?p,  and  the  outer  pressure  on  a  great  circle  whose  radius 
is  x-\-dx  is  n(x  -f-  dx)\p  +  dp).  The  difference  of  these  is 
equal  to  the  resistance  of  the  elementary  annulus,  which  is 
Zitxdx.S,  or 

(x  +  dx)\p  +  dp)  -  x*p  =  *Sxdx\ 


ART.  148.  HOLLOW  SPHERES.  335 

and  this,  omitting  quantities  of  the  second  order,  reduces  to 

xdp  +  2pdx  =  2$dx, 
which  is  a  second  equation  between  5  and  /. 

Substituting  in  the  second  equation  the  value  of  5  from  the 
first,  and  integrating,  the  value  of  /  in  terms  of  x  is  found, 
and  thus  5  also  ;  the  results  are 

s=c'+^'    *  =  c>-2i?< 

in  which  C9  is  a  constant  of  integration.  To  find  the  values  of 
Cl  and  C%>  the  value  of/  becomes  —  /,  when  x  =  rlt  and  —  /, 
when  x  =  r%  ;  thus 


and  these  are  the  formulas  for  thick  hollow  spheres  deduced 
by  LAME.  It  is  seen  that  they  are  analogous  to  those  for 
thick  cylinders,  the  radii  being  cubed  instead  of  squared.  It 
is  also  seen  that  the  formula  for  5  is  the  important  one  and 
that  its  greatest  value  occurs  at  the  inner  surface  of  the 
sphere. 

For  the  common  case  of  inner  pressure,  let/t=o,  and  make 
x  =  rl9  then  the  greatest  tangential  stress  is 


.»*   *  <*»*•** 

which  is  the  common  formula  for  hollow  spheres.  This  gives 
the  apparent  tensile  unit-stress  at  the  inner  surface.  The  true 
unit-stress  at  the  inner  surface  is,  by  Art.  129, 

T  =  S  -  IS  4-  \i>  =  2r*  "*"  T*  -  ^ 
?l        r*  —  r'       3 

which  will  usually  be  found  to  be  less  than  S,. 

As  an  example,  let  a  cylinder  4  inches  in  inner  and  8  inches 
in  outer  radius  have  a  hemispherical  end  with  the  same  radii, 


336  PLATES,    SPHERES,    AND   COLUMNS.  CH.  XV. 

and  be  subject  to  an  inner  water-pressure  of  4000  pounds  per 
square  inch.  Then  the  apparent  tensile  stress  on  the  inner 
surface  of  the  hemisphere  is 

256  -4-  64 
5,  =  — ^-  .  4000  =  2  860  pounds  per  square  inch, 

while  the  true  tensile  stress  is 

i  024  +  64  4  ooo         , 

T.=  —       ~^^          ~  =  2640  pounds  per  square  inch, 
512  —  04        3 

which  shows  that  the  true  value  is  about  8  per  cent  less  than 
the  apparent. 

For  the  cylinder  the  apparent  and  true  tensile  unit-stresses 
at  the  inner  surface  are,  from  Arts.  138  and  140, 

_  r;  4-  r'  4r,*  +  r,°S. 

•s'-^r^7/"       l='"^^7' 

which  L*ive  Sl  =  6  700  and  Tt  =  7  600  pounds  per  square 
inch,  so  that  the  true  stress  is  13  per  cent  greater  than  the 
apparent. 

If  the  end  of  the  cylinder  be  a  flat  plate  of  the  same  thick- 
ness  as  the  cylinder,  or  4  inches,  and  be  fixed  around  the  cir- 
cumference, the  true  stress  on  the  outer  side  is 

T  =  -f^  X  4  ooo  =  16  ooo  pounds  per  square  inch, 
which  is  6  times  as  great  as  for  the  hemisphere,  and  more 
than  double  the  greatest  stress  in  the  cylinder.  The  advan- 
tage of  hemispherical  ends  in  reducing  the  stresses  is  thus  seen 
to  be  very  great.  It  may  be  remarked,  in  conclusion,  that  the 
theory  of  internal  stress  in  cylinders  and  spheres  is  not  perfect, 
for  it  fails  to  give  the  same  results  for  the  common  surface  of 
junction  of  a  cylinder  and  hemisphere. 

Prob.  198.  A  hollow  sphere  is  to  be  subject  to  a  steam- 
pressure  of  600  pounds  per  square  inch,  its  inner  radius  being 
8  inches.  Find  its  thickness  so  that  the  greatest  stress  may 
be  i  ooo  pounds  per  square  inch. 


ART.  149.  EXPERIMENTS   ON   COLUMNS.  337 

Prob.  199.  Investigate  the  discrepancy  between  the  formu- 
las for  hollow  cylinders  and  hollow  spheres  for  the  following 
numerical  case.  A  hollow  cylinder  with  hollow  hemispherical 
ends,  the  inner  diameters  being  8  inches  and  the  outer 
diameters  12  inches,  is  subject  to  an  inner  water  pressure  of 
2400  pounds  per  square  inch.  Compute,  by  Art.  141  and  by 
this  article,  the  true  maximum  unit-stress  T  for  the  common 
plane  of  junction  of  cylinder  and  hemisphere. 

ART.  149.     EXPERIMENTS  ON  COLUMNS. 

It  is  impossible  to  present  here  even  a  summary  of  the  many 
experiments  that  have  been  made  to  determine  the  laws  of  re- 
sistance of  columns.  The  interesting  tests  made  by  CHRISTIE 
in  1883  f°r  tne  Pencoyd  Iron  Works  will  however  be  briefly 
described  on  account  of  their  great  value  and  completeness  as 
regards  wrought  iron  struts,  embracing  angle,  tee,  beam,  and 
channel  sections.  See  Transactions  of  the  American  Society 
of  Civil  Engineers,  April,  1884. 

The  ends  of  the  struts  were  arranged  in  different  methods; 
first  flat  ends  between  parallel  plates  to  which  the  specimen  was 
in  no  way  connected ;  second,  fixed  ends,  or  ends  rigidly 
clamped ;  third,  hinged  ends,  or  ends  fitted  to  hemispherical 
balls  and  sockets  or  cylindrical  pins  ;  fourth,  round  ends,  or  ends 
fitted  to  balls  resting  on  flat  plates. 

The  number  of  experiments  was  about  300,  of  which  about 
one-third  were  upon  angles,  and  one-third  upon  tees.  The 
quality  of  the  wrought  iron  was  about  as  follows :  elastic  limit 
32  ooo  pounds  per  square  inch.  Ultimate  tensile  strength 
49600  pounds  per  square  inch,  ultimate  elongation  18  per  cent 
in  8  inches.  The  length  of  the  specimens  varied  from  6  inches 
up  to  1 6  feet,  and  the  ratio  of  length  to  least  radius  of  gyra- 
tion varied  from  20  to  480.  Each  specimen  was  placed  in  a 
Fairbanks'  testing  machine  of  50  ooo  pounds  capacity  and  the 
power  applied  by  hand  through  a  system  of  gearing  to  two 


338 


PLATES,   SPHERES,    AND   COLUMNS. 


CH.  XV. 


rigidly  parallel  plates  between  which  the  specimen  was  placed 
in  a  vertical  position.  The  pressure  or  load  was  measured  on 
an  ordinary  scale  beam,  pivoted  on  knife  edges  and  carrying 
a  moving  weight  which  registered  the  pressure  automatically. 
At  each  increment  of  5  ooo  pounds,  the  lateral  deflection  of 
the  column  was  measured.  The  load  was  increased  until  failure 
occurred. 

The  following  are  the  combined  average  results  of  these  care- 


Length  divided  by 
Least  Radius  of 
Gyration. 

Flat  Ends. 

Fixed  Ends. 

Hinged  Ends. 

Round  Ends. 

20 

46000 

46000 

46000 

44000 

40 

4000O 

40000 

40000 

36500 

60 

36000 

36000 

36000 

30500 

80 

32000 

32000 

31500 

25000 

100 

29800 

30000 

28000 

20500 

1  20 

26300 

28000 

24300 

16  500 

,140 

23500 

25500 

21  000 

12800 

160 

20000 

23000 

16500 

95oo 

180 

16800 

20  OOO 

12800 

7500 

200 

14500 

17500 

I0800 

6000 

220 

12  700 

15000 

8800 

5000 

240 

II  200 

13000 

7500 

4300 

260 

9800 

II  OOO 

6500 

3800 

280 

8  500 

10  000 

5700 

3200 

300 

7200 

9000 

5000 

2800 

320 

6000 

8000 

4500 

2  5OO 

540 

5  ioo 

7000 

400O 

2  IOO 

360 

4300 

6500 

3500 

I  9OO 

380 

3500 

5  800 

3000 

I  700 

400 

3  ooo 

5  200 

.   2500 

I  500 

420 

2  500 

4800 

2300 

I  300 

440 

2200 

4300 

2  IOO 

460 

2000 

3  800 

I  900 

480 

I  900 

I  800 

ART.  150.  EULER'S  MODIFIED  FORMULA.  339 

fully  conducted  experiments.  The  first  column  gives  the 
values  of  l/r  and  the  other  columns  the  values  of  P/A,  the 
latter  being  the  ultimate  load  in  pounds  per  square  inch. 
From  the  results  it  will  be  seen  that  there  is  little  practical 
difference  between  the  strength  of  the  four  classes  when  the 
strut  is  short.  The  strength  of  the  long  columns  with  round 
ends  appears  to  be  about  3^  times  that  of  those  with  round 
ends. 

EULER'S  formula  fairly  represents  the  results  of  the  tests 
on  the  long  round-ended  columns.  Taking  E  =  25  ooo  ooo 
pounds  per  square  inch  for  wrought  iron,  and  «*  =  10,  Art. 

53  g^es, 

P  S 

-^  =  250000000^,, 

from  which  are  computed, 

for    l/r—    300,       340,       380,       400, 
P/A  =  2800,     2  200,      1700,      1400, 
while  the  experiments  give  the  values 

P/A  =  2800,     2100,      1700,      1300. 

Since  EULER'S  formula  is  deduced  under  the  laws  of  elasticity, 
it  must  be  concluded  that  the  elastic  limit  was  not  exceeded 
when  these  long  columns  failed  by  lateral  flexure. 

Prob.  200.  Plot  the  above  experiments  on  round-ended 
columns,  taking  P/A  as  abscissa  and  l/r  as  ordinate.  Also 
plot  on  the  same  sheet  EULER'S  curve  and  the  straight  line 
given  by  T.  H.  JOHNSON'S  formula. 

ART.  150.     EULER'S  MODIFIED  FORMULA. 

EULER'S  formula  for  columns  expresses  the  condition  of 
indifferent  equilibrium  or  that  state  which  borders  between 
stable  and  unstable  equilibrium.  In  all  cases  of  indifferent 
equilibrium  slight  causes  produce  marked  effects,  and  hence 
it  seems  important  to  inquire  whether  EULER'S  formula,  as 


340  PLATES,    SPHERES,   AND   COLUMNS.  CH.  XV. 

given  in  Art.  53,  represents  the  exact  relation  between  P/A 
and  l/r.  It  will  be  apparent  on  reflection  that,  while  the 
formula  contains  but  one  length  /,  there  are  really  three 
different  lengths  that  should  be  taken  into  consideration. 
Let  /  represent  the  length  of  the  straight  column  in  its  un- 
strained state,  /,  the  length  of  the  straight  column  after  com- 
pression by  the  unit-stress  P/A,  and  /8  the  chord  of  the 
deflected  curve  after  lateral  bending. 

Referring  now  to  formula  (5)  of  Art.  33,  it  is  seen  that  this 
does  not  include  the  effect  of  the  longitudinal  compression 
P/A.  To  introduce  this  let  5  be  the  total  unit-stress  pro- 
duced by  both  flexure  and  compression  ;  then  in  the  demon- 
stration the  first  four  formulas  will  be  thus  modified  : 


dl   .*«£    S~P'A     E    S^P/A     M 
di,~\>     ~7~     ~  R'        c       ~  r 

and  from  these  there  results, 


dx*~  Eldl,  "~  Ell? 

which  is  the  correct  differential  equation  of  the  elastic  curve 
for  a  body  under  combined  flexure  and  compression. 

Passing  now  to  EULER'S  deduction  in  Art.  53,  Mis  replaced 
by  —  Py,  and  the  equation  of  the  elastic  curve  for  a  round- 
ended  column  is 

A      ' 

y----A  sin 

Here  y  =  o  when  x  =  79;   hence  by  the  same  reasoning  as 
before, 

P=n>EI±     or     5=*'^ 

is  the  exact  condition  for  the  state  of  indifferent  equilibrium. 

To  complete  the  investigation  /,  and  /a  are  to  be  expressed 

in   terms  of  /.     Now  /  —  /,   is  the  deformation  due  to  the 

longitudinal  compression  P/A,  and  hence  from  the  fundamen- 


ART.  150.  EULER'S  MODIFIED  FORMULA.  341 

tal  definition  of  the  coefficient  of  elasticity  (Art.  4), 


which  gives  the  length  of  the  straight  column  after  longitu- 
dinal compression. 

To  find  /,  it  is  plain  that  /,  —  /,  is  the  fall  of  the  end  of  the 
column  due  to  the  lateral  flexure  and  that  P(/t  —  /,)  is  the 
work  done  in  this  fall.  This  external  work  must  equal  the 
internal  work  of  the  flexural  stresses.  From  Art.  109, 


JTs_  _  _ 

''  ~~  2EI      ~~  2EI  s 


and  integrating  this  between  the  limits  x  =  o  and  x  =  /,  ,  the 
internal  work  K  is  found  to  be  P*d*lJ^EI.  Thus,  equating 
the  two  works, 


and  from  this,  in  connection  with  the  above  value  of  /,  , 

i-P/AE    , 

~  i       P4'Ar** 


which  gives  the  length  of  the  chord  of  the  deflected  curve. 

The  quantities  P/AE  and  PA*/^Ar*  are  small  in  comparison 
with  unity,  and  hence  their  squares  and  their  product  may  be 
neglected;  also  the  reciprocal  of  I  —  P/AE  may  be  taken  as 
i  -f-  P/AE.  Introducing  the  values  of  /,  and  /,  into  the  above 
expression  for  P  it  reduces  to 

P  -r*  P          P 


and  since  n*Er*/l*  is  a  close  approximation  to  the  value  of 

P/A  it  may  replace  the  latter  in  the  parenthesis,  giving  finally 

P 


which  is  EULER'S  modifie'd  formula  for  round-ended  columns, 
By  writing  2\n*  instead  of  nra  it  applies  to  a  column  with  one 


342  PLATES,   SPHERES,    AND  COLUMNS.  CH.  XV. 

end  round  and  the  other  fixed,  and  by  writing  4^a  instead  of 
7ra  it  applies  to  a  column  with  both  ends  fixed. 

A  modification  of  EULER'S  formula  by  use  of  the  /t  and  / 
was  given  by  WlNCKLER  in  1867,  by  GRASHOF  in  1878, 
and  by  PRICHARD  in  1897  (see  Engineering  News,  May  6, 
1897).  The  formula  as  given  by  them  can  be  obtained  from 
the  above  by  making  /2  =  /,  or  by  making  A  =  o.  Since  the 
quantities  TrV2//2  and  ifA/T  are  small  compared  with  unity, 
EULER'S  modified  formula  gives  numerical  results  but  little 
larger  than  the  usual  ones.  It  shows,  however,  that  P  really 
increases  slightly  as  A  increases,  and  that  hence  A  is  in  strict- 
ness not  wholly  indeterminate,  as  the  common  theory  teaches. 

Prob.  201.  Show  that  the  fall  of  the  load  P,  due  to  direct 
compression  of  the  column,  equals  the  fall  due  to  the  lateral 
flexure  when  A  =  2r. 

ART.  151.     THE  DEFLECTION  OF  COLUMNS. 

An  ideal  column  always  remains  straight  under  the  action 
of  an  axial  load  W,  provided  that  this  load  is  less  than  the 
value  of  P  given  by  EULER'S  formula.  If  a  slight  lateral 
force  be  applied  it  will  deflect,  recovering  its  straight  condi- 
tion, however,  as  soon  as  this  force  is  removed.  Under  the 
action  of  such  a  slight  lateral  force  there  is  a  definite  relation 
between  the  deflection  A  and  the  maximum  unit-stress  5  on 
the  concave  side.  This  relation,  as  given  by  Art.  55,  is 

s 


and  it  shows  that  5  increases  with  A. 

Now  if  ^becomes  so  great  that  the  column  does  not  spring 
back,  but  remains  in  indifferent  equilibrium,  its  value  is  the  P 
given  by  EULER'S  formula,  and 

SI' 


Here  5  is  indeterminate,  and  the  column   may  remain  in  in- 


ART.   151.  DEFLECTION   OF  COLUMNS.  343 

different  equilibrium  with  many  different  values  of  A.  How- 
ever, if  A  be  so  great  that  5  becomes  equal  to  the  elastic 
limit  Se,  failure  at  once  follows,  and  hence  the  greatest  possi- 
ble deflection  is  found  by  using  Se  for  5  in  the  last  equation. 
When  a  steadily  increasing  load  W  is  applied  to  an  actual 
column  it  may,  on  account  of  being  non-homogeneous  or  not 
perfectly  straight,  begin  to  bend  without  the  application  of 
any  lateral  force  long  before  W  reaches  the  value  P  given  by 
EULER'S  formula.  Suppose,  for  instance,  that  the  column  is 
slightly  crooked  so  that  it  has  an  initial  deflection  6.  The 
actual  deflection  A  will  now  be  measured,  not  from  the  axis 
of  ordinates,  but  from  the  original  position  of  the  axis  of  the 
column.  Then  for  an  ideal  column  with  the  deflection  A  the 
bending  moment  is  PA,  while  for  the  actual  column  the  bend- 
ing moment  is  W(d  +  J),  and  equating  these  there  is  found, 

W 
*  =  p—W*< 

Thus  here  the  actual  deflection  increases  with  W,  and  no  in- 
determinateness  occurs.  This  case  in  fact  is  closely  analogous 
with  that  of  a  column  when  the  load  is  eccentric  with  respect 
to  the  axis,  the  deflection  A  being  always  perfectly  determi- 
nate (Art.  62).  Even  if  d  be  very  small  A  may  become  a 
considerable  quantity,  and  as  it  increases  the  unit-stress  5  also 
increases,  failure  being  practically  complete  as  soon  as  .S 
reaches  the  elastic  limit. 

These  conclusions,  it  will  be  seen,  are  essentially  the  same 
as  those  already  derived  in  Art.  62.  For  a  detailed  discussion 
of  columns  based  on  these  considerations  see  an  article  by 
MONTCRIEFF  in  Proceedings  of  American  Society  of  Civil 
Engineers  for  March,  1900,  or  in  its  Transactions,  Vol.  XLIII, 
1900. 


344  APPENDIX. 

APPENDIX. 

INTERNATIONAL  ASSOCIATION  FOR  TESTING  MATERIALS. 

In  1882  a  number  of  German  professors  of  engineering  met 
at  Munich  and  discussed  the  question  as  to  how  uniformity 
in  the  methods  of  testing  materials  might  be  promoted. 
Formal  conferences  were  held  in  other  German  cities  in 
1884,  1886,  1888,  and  1893,  at  which  engineers  frpm  other 
Europ'ean  countries  were  present.  The  reports  of  these 
conferences  attracted  wide  attention  and  the  movement 
assumed  an  international  character.  , 

In  1895  the  fifth  conference  met  at  Zurich,  all  European 
countries,  except  Turkey,  being  represented,  as  also  the 
United  States.  At  this  meeting  the  International  Associa- 
tion for  Testing  Materials  was  formally  organized,  its  object 
being  "  the  development  and  unification  of  standard  methods 
of  testing  for  the  determination  of  the  properties  of  materials 
of  construction  and  of  other  materials,  and  also  the  perfection 
of  apparatus  for  that  purpose."  This  may  be  called  the  first 
congress  of  the  Association.  The  second  congress  met  at 
Stockholm  in  1897,  there  being  present  361  members,  repre- 
senting eighteen  countries.  The  third  congress  was  intended 
to  be  held  at  Paris  in  1900,  but  this  was  omitted  on  account 
of  the  Exposition  Congress  on  the  same  subject  which  had 
been  organized  by  the  French  government. 

In  1899  the  total  number  of  members  of  the  International 
Association  was  about  I  700,  of  which  393  are  credited  to 
Russia,  384  to  Germany,  213  to  Austria,  140  to  the  United 
States,  87  to  England,  82  to  Switzerland,  77  to  France,  60 
to  Sweden,  42  to  Holland,  41  to  Norway,  and  the  remainder 
to  twelve  other  countries. 


VELOCITY   OF   STRESS.  345 

The  official  organ  of  the  Association  is  the  journal  Bau- 
materialienkunde,  published  semi-monthly  at  Stuttgart,  Ger- 
many. The  American  Section  of  the  Association,  organized 
in  1898,  publishes  occasional  Bulletins  containing  its  papers 
and  proceedings.  Bulletin  No.  4,  published  in  September, 
1899,  contains  an  address  on  the  work  of  the  Association  by 
the  Chairman  of  the  American  Section,  delivered  at  the  sec- 
ond annual  meeting  of  the  Section,  in  which  may  be  found 
detailed  information  regarding  the  work  and  plans  of  the 
Association.  Bulletin  No.  5  contains  a  valuable  report  on  the 
present  state  of  knowledge  regarding  impact  and  impact  tests. 

The  technical  work  of  the  Association  is  done  by  Interna- 
tional Committees  which  study  definite  problems  and  make 
reports  on  them  to  the  congresses.  At  the  congress  to  be  held 
in  1901  or  1902  it  is  expected  that  many  of  these  committees 
will  make  valuable  reports.  The  report  of  the  American 
branch  of  the  Committee  on  International  Specifications  for 
Testing  Iron  and  Steel  may  be  found  in  Bulletins  Nos.  8-18 
of  the  American  Section,  issued  in  May,  1900.  These  are 
accompanied  by  tables  showing  the  requirements  of  American 
manufacturers  and  consumers  regarding  the  chemical  compo- 
sition and  physical  properties  of  wrought  iron  and  various 
grades  of  steel,  and  they  also  contain  standard  specifications 
proposed  by  the  Committee. 

VELOCITY  OF  STRESS. 

When  an  external  force  is  suddenly  applied  to  a  body  the 
stresses  produced  are  not  instantaneously  generated,  but  are 
propagated  by  a  wave-like  motion  through  the  mass.  Hence 
there  is  a  velocity  of  transmission  of  stress  which  wiL  be 
shown  to  depend  upon  the  stiffness  and  density  of  the  mate- 
rial. In  fact,  a  sudden  stress  is  propagated  through  a  body 
in  the  same  manner  as  sound  is  propagated  through  the  air. 


346  APPENDIX. 

Let  v  be  the  velocity  with  which  stress  is  transmitted  in  a 
body  whose  coefficient  of  elasticity  is  E,  and  whose  weight  per 
cubic  unit  is  w  at  a  place  where  the  acceleration  of  gravity  is 
g.  It  is  required  to  find  v  in  terms  of  E,  wy  and  g. 

If  F  be  a  force  which  acting  continuously  for  one  second 
produces  the  velocity  #,  and  if  a  body  of  the  weight  W  ac- 
quires under  the  action  of  gravity  the  velocity^  in  one  second, 
then  the  forces  are  proportional  to  the  accelerations  that  they 
produce,  or 

# 
Fg  =  Wu,        whence         F  =  W—, 

o 

which  is  one  of  the  well-known  formulas  of  mechanics. 

Now  let  a  unit-stress  5  be  applied  to  the  end  of  a  bar,  pro- 
ducing the  unit-elongation  s  upon  the  first  element  of  its 
length.  The  elongation  of  the  first  element  transmits  the 
stress  to  the  second  element,  and  this  in  turn  produces  an 
elongation  of  the  second  element,  and  so  on.  At  the  end  of 
one  second  of  time  the  length  v  is  stressed,  and  the  total 
elongation  in  that  length  will  be  sv.  Thus  in  one  second  the 
center  of  gravity  of  the  bar  is  moved  the  distance  %sv,  and  its 
velocity  u  at  the  end  of  the  second  is  sv.  Now  referring 
to  the  formula  Fg  =  Wu,  the  value  of  F  is  S,  which  is  equal  to 
Es,  the  value  of  W  is  wv,  since  the  cross-section  considered  is 
unity,  and  hence  Esg  =  wv .  sv,  whence 


v  = 


which  is  the  formula  for  the  velocity  of  wave  propagation  in 
elastic  media  first  deduced  by  NEWTON. 

Taking  for  E  and  w  their  mean  values  given  in  Art.  80,  and 
also  g  as  32.16  feet  per  second  per  second,  since  the  unit-weights 
w  are  given  for  the  surface  of  the  earth,  the  mean  values  of  the 
velocity  of  transmission  of  stress  for  different  materials  are 
found  to  be  as  follows : 


ADVANCED   PROBLEMS.  347 

For  timber,  v  =  13  200  feet  per  second, 

For  stone,  v  =  13  200  feet  per  second, 

For  cast  iron,  v  •=  12  400  feet  per  second, 
For  wrought  iron,  v  =  15  500  feet  per  second, 
For  steel,  v  =  17  200  feet  per  second. 

In  making  this  computation  E  must  be  taken  in  pounds  per 
square  foot,  since  both  w  and  g  are  expressed  in  terms  of  feet. 

The  velocity  of  sound,  light,  and  all  wave  propagations  in 
elastic  media  is  given  by  the  above  formula  for  v.  The  ratio 
of  w  to  g  is  a  constant  for  the  same  material  at  any  point  in 
space,  and  it  expresses  the  density,  while  E  is  an  index  of  stiff- 
ness. The  ether  that  transmits  waves  of  light  must  be  lighter 
than  air  and  stiffer  than  steel,  in  order  that  v  may  be  the  high 
value  found  by  observation. 

Prob.  203.  What  time  is  required  for  sound  to  travel  a  dis- 
tance of  5  miles  in  water,  the  linear  unit-compression  for  water 
being  0.00005  ? 

ADVANCED  PROBLEMS. 

Many  questions  relating  to  flexure  and  torsion  have  not 
been  treated  in  the  preceding  pages,  as  their  discussion  prop- 
erly belongs  to  special  works  on  special  branches  of  applied 
mechanics.  A  few  of  these  are  here  noted  as  advanced  exer- 
cises that  may  be  assigned  by  teachers  as  prize  problems. 

Prob.  204.  Prove  that  the  maximum  bending  moment  caused 
by  two  equal  loads  rolling  over  a  simple  beam  occurs  at  the 
section  distant  ±a  from  the  middle,  a  being  the  distance  be- 
tween the  two  loads. 

Prob.  205.  Prove  that  the  maximum  bending  moment  at  a 
given  section  in  a  simple  beam,  due  to  a  given  system  of  mov- 
ing loads,  occurs  when  the  sum  of  those  on  the  left  of  the  sec- 
tion divided  by  the  distance  of  the  section  from  the  left  end 
equals  the  total  load  divided  by  the  length  of  the  span. 


348  APPENDIX. 

Prob.  206.  Prove  that  the  maximum  maximorum  bending 
moment  in  a  simple  beam,  due  to  a  given  system  of  moving 
loads,  occurs  at  a  section  so  located  that  the  distance  between 
it  and  the  center  of  gravity  of  the  loads  is  bisected  by  the 
middle  of  the  span. 

Prob.  207.  Let  a  helical  spring  consist  of  round  wire,  let  r 
be  the  radius  of  the  coil,  d  the  diameter  of  the  wire,  and  P  the 
tensile  and  compressive  load  upon  the  spring.  Show  that 

i6Pr 
S~'-~^d* 
is  the  shearing  or  torsional  unit-stress  in  the  wire. 

Prob.  208.  The  data  being  the  same  as  in  the  last  problem, 
and  n  being  the  number  of  coils,  show  that 


is  the  elongation  or  compression  of  the  spring. 

Prob.  209.  A  simple  beam  of  given  rectangular  cross-section 
carries  a  load  of  w  pounds  per  linear  unit  in  addition  to  its  own 
weight.  If  5  be  the  allowable  working  stress  show  that  the 
greatest  possible  length  of  the  beam  is 

/=    -  -     s* 


+  bdu)  ' 

where  u  is  the  weight  of  a  cubic  unit  of  the  material,  b  is  the 
breadth  and  d  the  depth  of  the  beam. 

Prob.  210.  If  the  flanges  of  an  I  beam  be  considered  to 
carry  all  the  bending  moment  and  the  web  all  the  vertical 
shear,  show  that  the  most  advantageous  proportions  are  such 
that  the  cross-section  of  the  flanges  equals  the  cross-section  of 
the  web. 

Prob.  211.  If  y  be  the  elongation  of  a  spring  or  bar  under 
longitudinal  impact,  prove  that 


is  the  time  of  one  oscillation. 


ADVANCED   PROBLEMS.  349 

Prob.  212.  Show  that  the  percentage  of  weight  saved  by 
using  a  hollow  instead  of  a  solid  shaft  is  -TTT  if  tnev  are  made 

of  equal  resilience,  n  being  the  ratio  of  the  outer  to  the  inner 
diameter  of  the  hollow  section. 

Prob.  213.  Show,  for  a  shaft  of  square  cross-section,  that 
the  formulas  for  investigation  and  design  are 

St=267Soo^-,,         d-- 

in  which  the  letters  have  the  same  meaning  as  in  Art.  68. 

Prob.  214.  A  load  P  is  supported  by  three  strings  of  equal 
size  lying  in  the  same  vertical  plane.  The  middle  string  is 
vertical  and  each  of  the  others  makes  an  angle  0  with  it.  If  S 
be  the  stress  on  the  middle  string  and  5,  the  stress  on  each  of 
the  others,  show  that 

5  -  P  S  -      P  COS>  ° 

"    I  +  2  COS'  0  '  l  ~  I  -f-  2  COS'  P 

(Note  :  To  solve  this  problem  the  condition  must  be  intro- 
duced that  the  internal  work  of  all  the  stresses  is  a  minimum.) 

Prob.  215.  A  load  P  is  supported  by  three  strings  of  equal 
size  lying  in  the  same  plane.  The  middle  string  is  vertical,  one 
string  makes  with  it  the  angle  6  on  one  side,  and  the  second 
string  makes  with  it  the  angle  0  on  the  other  side.  Find  the 
stresses  in  the  strings. 

Prob.  216.  A  circular  ring  of  radius  R  is  pulled  in  the  direc- 
tion of  a  diameter  by  two  tensile  forces  each  equal  to  P. 
Show  that  the  maximum  bending  moment  is  at  the  section 
where  P  is  applied,  and  that  its  value  is 

M-         PR* 

in  which  r  is  the  radius  of  gyration  of  the  cross-section  of  the 
ring.  Deduce  also  an  expression  for  the  maximum  negative 
bending  moment. 

Prob.  217.  A  continuous  beam  of  five  equal  spans  has  a  load 


350  APPENDIX. 

P  on  the  second  span  at  a  distance  kl  from  the  second  support. 
Show  that  the  reaction  of  the  first  support  is 


and  that  the  reaction  of  the  second  support  is 
J?,  =  ^(209  -  45^  -  38i 

Prob.  218.  Discuss  the  formula  tan20=//2z/  in  Art.  75 
and  show  that  it  represents  two  sets  of  shear  curves,  each 
being  at  right  angles  to  the  other.  Draw  the  two  sets  of 
curves  for  the  case  of  a  simple  beam  of  rectangular  cross- 
section. 

Prob.  219.  A  beam  of  two  equal  spans  has  a  joint  at  the 
middle  of  the  first  span  so  that  the  moment  there  is  always 
zero.  Find  the  reactions  due  to  a  load  P  on  the  first  span  ; 
(a)  when  k  is  less  than  £  ;  (b)  when  k  is  greater  than  \.  Draw 
for  each  case  the  diagrams  of  shears  and  moments. 

Prob.  220.  A  cantilever  bridge  has  three  spans,  the  length  of 
each  end  span  being  /  and  that  of  the  middle  span  m.  The 
middle  span  has  two  joints,  the  distance  of  each  from  the 
nearest  support  being  n.  Find  the  reactions  at  the  supports  ; 
(a)  when  the  load  P  is  on  an  end  span  ;  (b)  when  it  is  on  the 
middle  span  between  a  pier  and  the  nearest  joint  ;  (c)  when  it 
is  on  the  middle  span  between  the  two  joints. 

Prob.  221.  A  beam  is  fixed  at  the  ends  A  and  C,  and  loaded 
at  B  with  a  load  P  ;  the  end  C,  however,  being  free  to  deflect, 
while  B  and  C  are  kept  on  the  same  level.  Show  that  the  re- 
actions at  A  and  C  are 

4*  -3*     R__Pfr-sr 

*  -  p~^Jt'       R'-P  -J=W 

in  which  k  represents  the  ratio  of  BC  to  AC. 

Prob.  222.  A  continuous  beam  of  three  spans  has  each  end 
span  of  length  /  and  the  middle  span  of  length  «/.  Find  the 
reactions  due  to  a  load  P  in  an  end  span. 


ANSWERS  TO  PROBLEMS.          35 l 

ANSWERS  TO  PROBLEMS. 

Below  will  be  found  the  answers  to  about  nine-tenths  of  the 
problems  stated  in  the  preceding  pages,  the  number  of  the 
problem  being  in  parenthesis  and  the  answer  immediately  fol- 
lowing. It  has  been  thought  well  that  some  answers  should 
be  omitted  in  order  that  the  student  may  struggle  with  them 
to  ascertain  the  truth,  according  to  his  best  knowledge  of  the 
subject,  rather  than  to  make  his  numerical  results  agree  with 
given  figures.  However  satisfactory  it  may  be  to  the  student 
to  know  the  result  of  an  exercise  he  is  to. solve,  let  him  remem- 
ber that  after  commencement  day  the  answers  to  problems  will 
never  be  given. 

The  unit-stresses  to  be  employed  in  solutions  will  be,  unless 
otherwise  stated  in  the  problem,  uniformly  taken  from  the 
tables  given  in  the  text  and  in  Art.  80.  Considering  the  great 
variation  in  these  data  it  has  not  been  thought  best  to  carry 
the  numerical  answers  to  more  than  three  significant  figures, 
but  in  making  the  solution  four  significant  figures  should  be 
retained  through  the  work  in  order  that  the  third  may  be  correct 
in  the  final  result. 

Chapter  I.     (A)  7.2,  7.06,  and  86.4   square  inches.     (2)   173, 
34.7,  and  4.44  pounds.     (3)  55  100   pounds  per  square  inch. 
(4)  44700  pounds.     (5)   165000   pounds.      (6)    0.15    inches 
(7)  26250000  pounds  per  square   inch.     (8)    0.004    inches. 
(9)  About  4^  inches  X  4i  inches.     (10)  0.065  and  0.108  inches.    ', 
(12)   0.00153    inches.      (13)  52900    pounds    per   square   inch. 
(14)  849  pounds  per  square  inch.     (15)  About  1}  inches  in  di 
ameter.     (16)  9  for  AB  and  23  for  BC. 

Chapter  II.  (17)  0.88  inches  if/=  15.  (18)  I  170  pounds 
per  square  inch.  (19)  2  500  pounds  per  square  inch.  (20)  I  620 
pounds  per  square  inch.  (22)  2|  inches.  (23)  57  per  cent ; 
/=  7.7.  (24)  3^  inches;  about  0.72.  (25)  0.0032  inches. 


352  APPENDIX. 

Chapter  III.  (27)  2\  inches.  (29)  998  and  742  pounds. 
(3°)  +  800,  +  *  60,  and  —  180  at  I,  3,  and  5  feet  from  left  end. 
(31)  —  10,  —  40,  —  90,  —  40,  —  10  pound-feet.  (33)  Y=  140  and 
X  =  242  pounds.  (34)  2700  pounds.  (35)  X  —  —  Z  —  375 
pounds.  (36)  About  28.  (37)  4.20  inches.  (38)  c—  1.714 
inches,  7=  7.39  inches4.  (39)  -fabd*  and  %bd\  (41)  At  5.37 
feet  from  left  end  ;  M=  689  pound-feet.  (42)  No.  (43)  The 
bar  will  break.  (44)  294  pounds  per  linear  foot.  (45)  About 
6000  pounds.  (47)  8.87  inches.  (48)  0.0178  inches.  (49)  About 
610  pounds.  (51)  Ultimate  strengths  about  as  4  to  I,  while 
working  strengths  for  a  steady  load  are  about  as  3.7  to  i. 
(52)  3.0  and  1.5.  (53)  6  feet,  5  inches.  (54)  The  beam  will 
break.  (56)  5=5610  and  S'  =  3  170  pounds  per  square 
inch.  (57)  209  inches,  418  inches,  and  oo.  (58)  0.622  inches. 
(59)  J4  5°°  °°o  pounds  per  square  inch.  (62)  As  8  to  3  ;  as  64 

7      72 

to   9.    (63)  7^  inches.     (64)  0.243   inches.     (65)  x  —  6  ooo  —p 

for  the  first  case  ;   the  shear  at  supports  is  independent  of  x. 
(68)  0.72  inches. 

Chapter  IV.  (69)  The  diagrams  should  always  be  drawn  on 
cross-section  paper.  (70)  Ri  =  290.  (72)  £=0.366,  and 
£=0.577.  (73)  /=  2.828  m.  (74)  Max.  positive  M=  120 
pound-feet.  (76)  A  light  1  5-inch  beam  ;  a  heavy  12-inch  beam. 
(78)  0.0269  inches.  (80)  R^  —  R.  =  -&wl\  R,  =  R9  =  \\wl. 


(81)  —o°  25'  47".    (83)  -  =  7.2  which  requires  the  light  6-inch 

beam.     (84)  —-fowl.     (86)  n  =0.6095. 

Chapter  V.  (88)  9.42  inches.  (89)  2  inches.  (90)  205  ooo 
pounds.  (92)  69.7  tons.  (93)  5.05  inches.  (95)  r  =  0.86 
inches.  (96)  if.  (97)  2.35  and  24.  (98)  168  ooo  pounds. 
(99)  23  ooo  pounds.  (100)  13^  and  i6f  inches  square.  (104) 
Draw  Fig.  48  so  as  to  make  bq  =  O',  then  state  the  equation 
of  moments  and  reduce  it  by  the  relation  between  the  similar 
triangles. 


ANSWERS  TO   PROBLEMS.  353 

Chapter  VI.  (106)  30  pounds.  (107)  105  degrees.  (108) 
720,  270,  and  290  pound-inches.  (109)  I  876  pounds  per  square 
inch.  (no)  /=  0.0361^*  and  <:=  0.577^.  (in)  J  =  23.6 
and  £=3. 41.  (112)  I  680  pounds,  (i  13)  9  380000  pounds 
per  square  inch,  (i  14)  65  horse-power.  (115)9.7.  (116)2.65 
and  3.58  inches.  (118)6500.  ( 1 19)  As  V^n  to  3.  ( 1 20)  As 
100  to  1 06. 

Chapter  VII.  (122)  3720  pounds  per  square  inch.  vI23) 
8  100  pounds  per  square  inch.  (124)  The  light  9-inch  beam. 
(125)  Nearly  8  inches.  •  (126)  9  inches.  (127)  /  =  9  420, 
0  -  54°  13';  S==  7160,  0-9°  13'.  (129)  5.4.  (130)  2j 
inches.  (131)  S  =  5  660  and  Ss  =  205  pounds  per  square  inch. 
(132)  At  3  inches  from  neutral  surface  S  =  2OOO,  Sk  =  250, 
and  /  =  2  030  pounds  per  square  inch. 

Chapter  IX.  (139)  Theoretic  stress  is  3.3  per  cent  and 
theoretic  elongation  is  3.5  per  cent  greater  than  the  observed 
values.  (140)  1.34  horse-power.  (141)  122  foot-pounds. 
(142)  For  the  second  case  K=  S*Al/\$E  if  section  is  rect- 
angular. 

Chapter  X.    (144)  7  500  feet  and  3.75  feet.     (145)  log  x  — 

wbl 
l°g  b  +  °-4343  —p-y-    (146)  About  101  feet  per  second.    (148) 

Nearly  28000  pounds  per  square  inch.  (149)  64  rollers  2 
inches  in  diameter,  or  7  rollers  6  inches  in  diameter.  (150)  6. 
(151)  1 1  500  pounds  per  square  inch. 

Chapter  XI.  (152)  For  uniform  load  M  =  —  \wx*  and 
K  =  W*l*/4pEI.  (153)  Deflection  =  T\th  of  that  for  load  at 
middle.  (154)^  =  o.  1 74  inches,  5  —  4030  pounds  per  square 
inch.  (157)  1420  pounds,  (158)  About  1.4  inches. 
(160)  Over  600  miles  per  hour.  (161)  Wtl/2^EsA.  (166)  See 
Higher  Mathematics,'  Chap.  IV,  Art.  36. 

Chapter  XII.  (167)  £(6000)  —  *(—  2  ooo)  =  4000.  (168) 
About  12  horse-power.  (170)  £,  =  9380000,  then  find  e 


354  APPENDIX. 

from  formula  (25).  (171)  As  2  to  3.  (172)  As  100  to  307. 
(173)  8  bolts.  (174)  2TV  inches.  (175)  Bearing  unit-stress  — 
i  880  pounds  per  square  inch.  (176)  Shear  is  uniform  through- 
out, while  moment  is  zero  at  middle. 

Chapter  XIII.  (178)  0.042  and  0.014  per  cent.  (181)  S6°I9' 
with  axis  of  bolt.  (183)  See  Theory  of  Equations  in  Algebra. 
(184)  Greatest  tension  =  36.3,  compression  =  87.1  pounds  per 
square  inch.  (185)  Apparent  =  53.7,  true  =  61.7  pounds  per 
square  inch.  (186)  54°  44'  with  greater  and  3 5°  46'  with  lesser 
stress.  (187)  Maximum  true  compression  =  12900  pounds 
per  square  inch,  or  27  per  cent  more  than  the  apparent. 

Chapter  XIV.  (188)  Make  rl  =  o.  (189)  18000  pounds  per 
square  inch.  (190)  54000  pounds  per  square  inch.  (191) 
15900  pounds  per  square  inch.  (192)  15000  pounds  per 
square  inch.  (194)  Deduce  an  expression  for  pv  in  terms  of 
the  given  radii  and  Se ;  then  find  the  value  of  r9  which  renders 
pl  a  maximum. 

Chapter  XV.  (195)  See  the  books  mentioned  in  Art.  127. 
(196)  About  50  pounds  per  square  inch  with  factor  of  10.  (198) 
Thickness  =  1.6  inches.  (201)  See  the  demonstrations  in 
Arts.  55  and  108. 

Appendix.  (202)  See  Railroad  Gazette,  June  7,  1895.  (205) 
See  Roofs  and  Bridges,  Part  I.  (217)  See  London  Philosoph- 
ical Magazine,  September,  1875.  (218)  See  Engineering  News, 
August  I,  1895.  (221)  See  article  by  J.  L.  GREENLEAF  in 
Journal  of  Franklin  Institute  for  July,  1895. 

^DESCRIPTION  OF  TABLES. 

Tables  I,  II,  III,  and  IV,  in  Art.  80  (pages  163  and  164) 
give  mean  constants  of  the  elasticity  and  strength  of  the  prin- 
cipal materials  used  in  engineering.  Other  tables,  not  num- 
bered, are  noted  in  the  Table  of  Contents  (page  ix). 


DESCRIPTION   OF  TABLES.  355 

Table  V,  on  the  next  two  pages,  gives  four-place  logarithms 
of  numbers  which  will  be  found  very  useful  and  sufficiently 
accurate  for  all  computations  in  the  mechanics  of  materials. 

Table  VI  gives  four-place  squares  of  numbers  from  i.oo  to 
9.99,  the  arrangement  being  the  same  as  that  of  the  logarithmic 
table.  By  properly  moving  the  decimal  point  four-place 
squares  of  other  numbers  may  also  be  taken  out.  For  example, 
the  square  of  0.874  is  0.7639,  that  of  87.4  is  7*639,  and  that  of 
874  is  763  900,  correct  to  four  significant  figures. 

Table  VII  gives  four-place  areas  of  circles  for  diameters 
ranging  from  i.oo  to  9.99,  arranged  in  the  same  manner.  By 
properly  moving  the  decimal  point  four-place  areas  for  all  circles 
maybe  found.  For  instance,  if  the  diameter  is  4.175  inches, 
the  area  is  13.69  square  inches;  if  the  diameter  is  0.535  inches 
the  area  is  0.2248  square  inches  ;  if  the  diameter  is  12.3  feet,  the 
area  is  116.9  square  feet,  all  correct  to  four  significant  figures. 

Table  VIII  gives  weights  per  linear  foot  of  wrought-iron  bars 
both  square  and  round,  the  side  of  the  square  or  the  diameter 
of  the  circle  ranging  from  -J  to  lof  inches.  Approximate 
weights  of  bars  of  other  materials  may  be  derived  from  this 
table  by  the  following  rules : 

For  timber,       divide  by  12; 

For  brick,         divide  by  4  ; 

For  stone,         divide  by  3  ; 

For  cast  iron,  subtract  6  per  cent ; 

For  steel,          add  2  per  cent. 

For  example,  a  cast-iron  bar  6£  inches  square  and  8  feet  long 
weighs  8(157.6  —  0.06  X  157.6)  =  I  185  pounds.  In  like  man- 
ner a  steel  bar  2T3T  inches  in  diameter  and  4  feet  9  inches  in 
length  weighs  44(12.53  +0.02  X  12.53)  =  57'9  pounds. 


356 


APPENDIX 


TABLE   V.      COMMON    LOGARITHMS. 


n 

01234 

56789 

Diff. 

IO 

oooo  0043  0086  0128  0170 

0212   0253   0294   0334   0374 

42 

ii 

0414  0453  0492  0531  0569 

0607   0645   0682   0719   0755 

38 

12 

0792  0828  0864  0899  0934 

0969  1004  IO38  1072  1106 

35 

14 

1139  1173  1206  1239  1271 

1461  1492  1523  1553  1584 

!3°3  T335  !367  1399  '43° 
1614  1644  1673  J7Q3  1732 

32 
3° 

15 

1761  1790  .1818  1847  J875 

1903  1931  1959  1987  2014 

28 

16 

2O4I   2O68   2095   2122   2148 

2175   2201   2227   2253   2279 

27 

17 

2304   2330   2355   2380   2405 

2430  2455  2480  2504  2529 

25 

18 

2553   2577   266l   2625   2648 

2672  2695  27*8  2742  2765 

24 

T9 

2788   28lO   2833   2856   2878 

2900  2923  2945  2967  2989 

22 

20 

3010   3032   3054   3075   3096 

3IlS  3139  3J6o  3181  3201 

21 

21 

3222   3243   3263   3284   3304 

3324  3345  3365  3385  3404 

2O 

22 

3424  3444  3464  3483  3502 

3522  354i  3560  3579  3598 

19 

23 

3617  3636  3655  3674  3692 

37ii  3729  3747  3766  3784 

18 

24 

3802  3820  3838  3856  3874 

3892  3909  3927  3945  3962 

18 

25 

3979  3997  4014  4031  4048 

4065  4082  4099  4116  4133 

17 

26 

4150  4166  4183  4200  4216 

4232  4249  4265  4281  4298 

17 

27 

4314  4330  4346  4362  4378 

4393  4409  4425  4440  4456 

16 

28 

4472  4487  4502  4518  4533 

4548  4564  4579  4594  4609 

15 

29 

4624  4639  4654  4669  4683 

4698  4713  4728  4742  4757 

'5 

30 

4771  .4786  4800  4814  4829 

4843  4857  4871  4886  4900 

14 

4914  4928  4942  4955  4969 

4983  4997  5011  5024  5038 

14 

32 

5051  5065  5079  5092  5105 

5119  5132  5145  5159  5172 

13 

33 

5185  5198  5211  5224  5237 

5250  5263  5276  5289  5302 

13 

34 

53*5  5328  5340  5353  5366 

5378  5391  5403  5416  5428 

13 

35 

544i  5453  5465  5478  5490 

5502  5514  5527  5539  5551 

12 

36 

5563  5575  5587  5599  5^n 

5623  5635  5647  5658  5670 

12 

37 

5682  5694  5705  5717  5729 

5740  5752  5763  5775  5786 

12 

38 

5798  5809  5821  5832  5843 

5855  5866  5877  5888  5899 

II 

39 

5911  5922  5933  5944  5955 

5966  5977  5988  5999  6010 

II 

40 

6021  6031  6042  6053  6064 

6075  6085  6096  6107  6117 

II 

41 

6128  6138  6149  6160  6170 

6180  6191  6201  6212  6222 

II 

42 

6232  6243  6253  6263  6274 

6284  6294  6304  6314  6325 

IO 

43 
44 

6335  6345  6355  6365  6375 
6435  6444  6454  6464  6474 

6385  6395  6405  6415  6425 
6484  6493  65°3  65r3  6522 

10 
IO 

^!5 

6532  6542  6551  6561  6571 

6580  6590  6599  6609  6618 

10 

46 

6628  6637  6646  6656  6665 

6675  6684  6693  6702  6712 

9 

47 

6721  6730  6739  6749  6758 

6767  6776  6785  6794  6803 

9 

48 

6812  6821  6830  6839  6848 

6857  6866  6875  6884  6893 

9 

49 

6902  6911  6920  6928  6937 

6946  69^5  6964  6972  6981 

9 

5° 

6990  6998  7007  7016  7024 

7033  7042  7050  7059  7067 

9 

7076  7084  7093  7101  7110 

7118  7126  7135  7143  7152 

8 

52 

7160  7168  7177  7185  7193 

7202  7210  7218  7226  7235 

8 

53 

7243  7251  7259  7267  7275 

7284  7292  7300  7308  7316 

8 

54 

7324  7332  7340  7348  7356 

7364  7372  7380  7388  7396 

8 

n 

01234 

56789 

Diff. 

COMMON  LOGARITHMS. 


357 


TABLE   V.      COMMON    LOGARITHMS. 


n 

01234 

56789 

Diff. 

55 

7404  7412  7419  7427  7435 

7443  7451  7459  7466  7474 

8 

56 

7482  7490  7497  75°5  75  '3 
7559  7566  7574  7  582  7589 
7634  7642  7649  7657  7664 

7520  7528  7536  7543  7551 
7597  7604  7612  7619  7627 
7672  7679  7686  7694  7701 

59 

7709  7716  7723  7731  7738 

7745  7752  776o  7767  7774 

60 
61 

7782  7789  7796  7803  7810 
7853  7860  7868  7875  7882 

7818  7825  7832  7839  7846 
7889  7896  7903  7910  7917 

7 

62 

7924  7931  7938  7945  7952 

7959  7966  7973  798o  7987 

63 

7993  8000  8007  8014  8021 

8028  8035  8041  8048  8055 

• 

64 

8062  8069  8075  8082  8089 

8096  8102  8109  8116  8122 

65 

8129  8136  8142  8149  8156 

8162  8169  8176  8182  8189 

7 

66 

8195  8202  8209  8215  8222 

8228  8235  8241  8248  8254 

67 

8261  8267  8274  8280  8287 

8293  8299  8306  8312  8319 

68 

832?  8331  8338  8344  8351 

8357  8363  8370  8376  8382 

69 

8388  8395  8401  8407  8414 

8420  8426  8432  8439  8445 

70 

8451  8457  8463  8470  8476 

8482  8488  8494  8500  8506 

6 

7r 

8513  8519  8525  8531  8537 

8543  8549  8555  8561  8567 

72 
73 

8573  8579  8585  8591  8597 
8633  8639  8645  8651  8657 

8603  8609  8615  8621  8627 
8663  8669  8675  8681  8686 

74 

8692  8698  8704  8710  8716 

8722  8727  8733  8739  8745 

75 

8751  8756  8762  8768  8774 

8779  8785  8791  8797  8802 

6 

76 

8808  8814  8820  8825  8831 

8837  8842  8848  8854  8859 

77 

8865  8871  8876  8882  8887 

8893  8899  8904  8910  8915 

78 
79 

8921  8927  8932  8938  8943 
8976  8982  8987  8993  8998 

8949  8954  8960  8965  8971 
9004  9009  9015  9020  9025 

80 

9031  9036  9042  9047  9053 

9058  9063  9069  9074  9079 

5 

81 
82 

9085  9090  9096  9101  9106 
9i3»  9M3  9M9  9T54  9'59 

9112  9117  9122  9128  9r33 
9165  9170  9175  9180  9186 

83 

9191  9196  9201  9206  9212 

9217  9222  9227  9232  9238 

84 

9243  9248  9253  9258  9263 

9269  9274  9279  9284  9289 

85 

9294  9299  9304  9309  9315 

9320  9325  9310  9375  9340 

5 

86 

9345  9350  9355  936o  9365 

9370  9375  938o  93g5  9390 

87 

9395  94oo  9405  94:0  9415 

9420  9425  9430  9435  9440 

88 

9445  9450  9455  946  >  9465 

9469  9474  9479  9484  9489 

89 

9494  9499  95°4  95°9  95  '  3 

95  l8  9523  9528  9533  953s 

90 

9542  9547  9552  9557  9562 

9566  9571  9576  9581  9586 

5 

91 

9590  9595  96oo  9605  9609 

9614  9619  9624  9628  9633 

92 

9638  9643  9647  9652  9657 

9661  9666  9671  9675  9680 

93 

9685  9689  9694  9699  9703 

9708  9713  9717  9722  9727 

94 

973  i  9736  9741  9745  975° 

9754  9759  9763  9768  9773 

95 

9777  9782  9786  9791  9795 

9800  9805  9809  9814  9818 

4 

96 

9823  9827  9832  9836  9841 

9845  9850  9854  9859  9863 

97 

9868  9872  9877  9881  9886 

9890  9894  9899  9907  9908 

98 

9912  9917  9921  9926  9930 

9934  9939  9943  994»  9952 

99 

9956  9961  9965  9969  9974 

9978  9983  9987  9991  9996 

n 

01234 

5    6    7    8    9 

Diff. 

358 


APPENDIX. 


TABLE   VI.       SQUARES    OF   NUMBERS. 


7Z. 

01234 

56789 

Diff. 

.O 

i.ooo    1.020    1.040    1.061    1.082 

1.103    I-I24    I-I45    1.166    i.iSS 

22 

.1 

1.210    1.232    1.254    1.277    I-3°° 

1.323    1.346    1.369    1.392    1.416 

24 

.2 

1.440   1.464   1.488   1.513   1.538 

1.563    1.588    1.613    1.638    1.664 

26 

•3 

1.690    1.716    1.742    1.769    1.796 

1.823    1.850    1.877    1.904    1.932 

28 

-4 

1.960    1.988    2.016    2.045    2.074 

2.IO3     2.132     2.l6l      2.I9O     2.22O 

3° 

i 

2.250    2.280    2.310    2.341    2.372 
2.560    2.592    2.624    2.657    2.690 

2.403     2.434     2.465     2.496     2.528 
2.723     2.756     2.789     2.822     2.856 

32 
34 

•7 

2.890   2.924    2.958    2.993    3-028 

3.063     3.098     3.133     3.168     3.204 

36 

1.8 

3.240  3.276  3.312  3.349  3.386 

3.423     3.460     3.497     3.534     3.572 

1.9 

3.610  3.648  3.686  3.725  3.764 

3.803     3.842     3.881      3.920     3.960 

40 

2.O 

4.000   4.040   4.080   4.121    4.162 

4.203     4.244     4.285     4.326     4.368 

42 

2.1 

4.410  4.452  4.494  4.537  4.580 

4.623     4.666     4.709     4.752     4.796 

44 

2.2 

4.840   4.884   4.928    4.973    5.018 

5.063     5.108     5.153     5.198     5.244 

46 

2-3 

2.4 

5-290  5-336  5.382  5.429  5.476 
5.760  5.808  5.856  5.905  5.954 

5-523     5-570     5.617      5.664     5.712 

6.003    6.052    6.  i  oi    6.150   6.200 

48 
5° 

2-5 

6.250   6.300   6.350   6.401    6.452 

6.503    6.554    6.605    6.656   6.708 

52 

2.6 

6.760   6.8  1  2    6.864   6.917    6.970 

7.023  7.076  7.129  7.182  7.236 

54 

2.7 

7-290   7-344   7-398   7-453   7-5°8 

7.563  7.618  7.673  7.728  7.784 

56 

2.8 

7.840   7.896   7.952   8.009   8.066 

8.123  8.180  8.237  8.294  8.352 

2.9 

8.410   8.468   8.526   8.585   8.644 

8.703  8.762  8.821  8.880  8.940 

60 

3-° 

9.000   9.060   9.120   9.181    9.242 

9-303   9-364   9-425   9-4S6   9.548 

62 

3.1 

9.610   9.672    9.734   9.797    9.860 

9.923   9.986    10.05    I0-ir    IO-T8 

6 

3-2 

10.24    10.30    10.37    10.43    IO-5° 

10.56    10.63    10.69    10.76    10.82 

7 

3-3 

10.89    10.96    11.02    11.09    ii-i6 

11.22    11.29    IT-36    1142    11.49 

7 

34 

11.56    11.63    II-7°    JI-7°   ^1-83 

11.90     11.97      12.04     I2.II      12.  l8 

7 

3-5 

12.25    12.32    12.39    12.46    12.53 

1  2.60    12.67    I274    12.82    12.89 

7 

3-6 

12.96    13-03    13.10    13.18    13.25 

!3-32    1340    1347    13-54    13-62 

7 

3-7 

13.69    13.76    13.84    13.91    13.99 

14.06    14.14    14.21    14.29    14.36 

8 

3-8 

14.44    M-S2    H-59    M-67    14-75 

14.82    14.90    14.98    15.05    15.13 

8 

3-9 

15.21    15.29    15.37    15.44    15.52 

15.60    15.68    15.76    15.84    15.92 

8 

4.0 

16.00    16.08    16.16    16.24    16.32 

16.40    16.48    16.56    16.65    J6-73 

8 

4.1 

16.81    16.89    J6-97    17.06    17.14 

17.22    17.31    17.39    17.47    17-56 

8 

4.2 

17.64    17.72    17.81    17.89    17.98 

1  8.06    18.15    18.23    l8-32    18.40 

9 

4-3 

18.49    18.58    18.66    18.75    l8-84 

18.92    19.01    19.10    19.18    19.27 

9 

44 

19.36    19.45    19.54    19.62    19.71 

19.80    19.89    19.98    20.07    20.16 

9 

4-5 

20.25    20.34    20.43    20.52    20.61 

20.70    20.79    20.88    20.98    21.07 

9 

4.6 

21.  16   21.25    21.34    21.44    21.53 

21.62    21.72    21.81    21.90   22.00 

9 

4-7 

22.09   22.18    22.28    22.37    22.47 

22.56    22.66   22.75    22.85    22.94 

10 

4.8 

23.04   23.14    23.23    23.33    23.43 

23.52  23.62  23.72  23.81   23.91 

IO 

4-9 

24.01    24.11    24.21    24.30    24.40 

24.50    24.60    24.70    24.80    24.90 

IO 

5-o 
5-1 

25.00   25.10   25.20    25.30    25.40 
26.01    26.11    26.21    26.32    26.42 

25.50    25.60    25.70   25.81    25.91 
26.52   26.63   26.73   26.83   26.94 

IO 
IO 

5-2 

27.04    27.14    27.25    27.35    27.46 

27.56  27.67  27.77  27.88  27.98 

II 

5-3 

28.09   28.20    28.30   28.41    28.52 

28.62    28.73    28.84    28.94    29.05 

II 

5-4 

29.16   29.27    29.38    29.48    29.59 

29.70    29.81    29.92    30.03    30.14 

II 

n. 

01234 

56789 

Diff. 

SQUARES   OF    NUMBERS. 


359 


TABLE  VI.      SQUARES    OF   NUMBERS 


n. 

01234 

56789 

Diff. 

5-5 

30.25    30.36   30.47    30.58    30.69 

30.80   30.91    31.02    31.14   31.25 

I! 

5.6 

31.36   31.47    31.58    31.70   31.81 

31.92    32.04    32.15    32.26   32.38 

II 

5-7 

32.49   32.60   32.72    32.83    32.95 

33.06   33.18    33.29   33.41    33.52 

5.8 

33.64    33.76   33.87    33.99   34.  rr 

34.22    34.34    34.46   34.57    34.69 

5-9 

34-8i    34.93    35.05    35.16   35.28 

35-40   35.52    35.64   35.76   35.88 

6.0 

36.00    36.12    36.24    36.36   36.48 

36.60   36.72    36.84   36.97    37.09 

6.1 

37-21    37.33    37.45    37.38    37.70 

37.82    37.95    38.07    38.19   38.32 

6.2 

38.44    38.56   38.69   38.81    38.94 

39.06   39.19   39.31    39.44    39.56 

13 

6-3 

39-69   39-82    39.94   40.07    40.20 

40.32    40.45    40.38   40.70   40.83 

13 

6.4 

40.96   41.09   41.22    41.34   41.47 

41.60   41.73   41.86   41.99   42.12 

U 

6-5 

6.6 

42.25   42.38    42.51    42.64   42.77 
43-56   43-69   43-82    43-96   44-09 

42.90   43-03   43-  '6  43-3°   43-43 
44.22    44.36   44.49   44.62    44.76 

13 
U 

6.7 

44.89   45.02    45.16   45.29   45.43 

45.56   45.70   45-83   45-97    46.10 

U 

6.8 

46.24   46.38    46.51    46.65   46.79 

4692    47.06   47.20   47.33   47.47 

14 

6.9 

47.61    47.75   47-89    48.02   48.16 

48.30   48.44    48.58   48.72    48.86 

14 

70 

49.00   49.14    49.28   49.42    49.56 

49.70   49.84   49.98    50.13    5027 

U 

7-1 

50.41    50.55    50.69    50.84    50.98 

51.12    51.27    51.41    51.55    51.70 

M 

7.2 

51.84    51.98    52.13    52.27    52.42 

52.56    52.71    52.85    53.00    53.14 

15 

7-3 

53-29    53-44    53-58    53.73    53.88 

54.02    54.17    54.32    54.46    54.61 

15 

7-4 

54.76    54-91    55-o6   55-2o   55.35 

55-50    55.65    55.80    55.95    56.10 

15 

# 
ft 

56.25    56.40    56.55    56.70   56.85 
57-76   57.91    58.06    58.22    58.37 
59.29   59.44    59.60    59.75    59.91 
60.84   61.00   61.15   61.31    61.47 

C7.oo    57.15    57.30    57.46    57.61 
58.52    58.68    58.83    58.98    59.14 
60.06   60.22    60.37    60.53   60.68 
61.62    61.78    61.94   62.09   62.25 

IS 
15 

1  6 
16 

79  . 

62.41    62.57    62.73   62.88   63.04 

63.20   63.36   63.52    63.68   63.84 

16 

8.0 
8.1 

64.00  64.16  64.32   64.48   64.64 
65.61    65.77    65.93    66.  10   66.26 

64.80   64.96   65.12    65.29   65.45 
66.42    66.59   66.75   66.91    67.08 

16 
16 

8.2 

67.24   67.40   67.57    6773    67.90 

68.06   68.23    68.39   68.56   68.72 

17 

8-3 

68.89   69.06   69.22    69.39   69.56 

69.72    69.89   70.06   70.22    70.39 

17 

8.4 

70.56   70.73    70.90   71.06   71.23 

71.40   71.57    71.74    71.91    72.08 

17 

& 

72.25    72.42    72.59   72.76   72.93 

73-10   73-27    73-44   73-62    73-79 

17 

8.6 
8.7 

73-96   74-13   743°   74-48    74.65 
75.69   75.86   76.04   76.21.  76.39 

74.82    75.00   75.17    75.34   75-52 
76.56   76.74    76.91    77.09   77.26 

11 

8.8 
8.9 

77.44   77.62    77.79   77.97    78.15 
79.21    79.39   79.57    79.74    79.92 

78.32    78.50   78.68    78.85    79.03 
80.  10  80.28   80.46  80.64  80.82 

18 
18 

9.0 

8  1.  oo   81.18   81.36   81.54   81.72 

81.90  82.08  82.26  82.45   82.63 

18 

9.1 

82.81    82.99   83.17    83.36  83.54 

83.72   83.91    84.09  84.27   84.46 

18 

9.2 

84.64   84.82    85.01    85.19   85.38 

85.56  85.75  85.93  86.12   86.30 

i9 

9-3 

86.49   86.68    86.86   87.05   87.24 

87.42   87.61    87.80  87.98  88.17 

19 

94 

88.36   88.55   88.74   88.92   89.11 

89.30  89.49  89.68   89.87   90.06 

19 

9-5 

90.25   90.44   90.63   90.82    91.01 

91.20  91.39  91.58   91.78  91.97 

19 

9.6 

92.16   92.35   92.54   92.74   92.93 

93-  i  2    93.32    93.51    93.70   93.90 

19 

9-7 

94.09   94.28   94.48   94.67    94.87 

95.06   95.26   95.45   95.65   95.84 

20 

9.8 

96.04   96.24    96.43   96.63   96.83 

97.02   97.22    97.42    97.61    97.81 

20 

9-9 

98.01    98.21    98.41    98.60  98.80 

99.00   99.20   99.40   99.60   99.80 

20 

n. 

01234 

56789 

Diff. 

APPENDIX. 


TABLE   VII.       AREAS    OF   CIRCLES. 


d 

01234 

56789 

Diff. 

.0 

.7854  .8012  .8171  .8332  .8495 

.8659  .8825  .8992  .9161  .9331 

.1 

.9503    .9677    .9852    1.003    1.  021 

1.039    L057    1-075    1.094    1.  112 

.2 

1.131   1.150  1.169  1.188  1.208 

1.227    1.247    1.267    1.287    1.307 

19 

-3 

1.327  1.348  1.368  1.389  1.410 

1.431    1.453   1.474  1-496  I.5I7 

21 

•4 

1.539  I-561  I-584  i.  606  1.629 

1.651    1.674  1-697  1.720  1.744 

22 

•  5 

1.767  1.791  1.815  1.839  1-863 

1.887   1.911   1.936  1.961   1.986 

24 

.6 

2.0II    2.036    2.061    2.087    2.  112 

2.158  2.164  2.190  2.217  2.243 

26 

•  7 

2.270  2.297  2.324  2.351  2.378 

2.405  2.433  2.461   2.488  2.516 

27 

.8 

2-545  2.573  2.602  2.630  2.659 

2.638  2.717  2.746  2.776  2.806 

29 

•9 

2.835  2.865  2.895  2.926  2.956 

2.986  3.017  3.048  3.079  3.110 

30 

2.0 

3.142  3.173  3.205  3.237  3.269 

3-301   3.333  3.365  3.398  3.431 

32 

2.1 

3.464  3.497  3.530  3.563  3.597 

3.631   3.664  3.698  3.733  3.767 

34 

2.2 

3.801  3.836  3.871  3.906  3.941 

3.976  4.012  4.047  4.083  4.119 

35 

2-3 

4.155  4.191  4.227  4.264  4.301 

4-337  4-374  4-412  4.449  4.486. 

36 

2.4 

4.524  4.562  4.600  4.638  4.676 

4.7M  4-753  4-792  4-831  4-8/0 

38 

2-5 

4.909  4.948  4.988  5.027  5.067 

5.107  5.147  5.187  5  228  5.269 

40 

2.6 

5.309  5.350  5.391  5.433  5.474 

5-5I5  5-557  5-599  5-641  5.683 

4i 

2.7 

5.726  5.768  5.811  5.853  5.896 

5.940  5.983  6.026  6.070  6.114 

43 

2.8 

6.158    6.202    6.246    6.290   6.335 

6.379  6.424  6.469  6.514  6.560 

44 

2.9 

6.605  6.651  6.697  6.743  6.789 

6.835  6.881  6.928  6.975  7.022 

46 

3-0 

7.069  7.116  7.163  7.211  7.258 

7.306  7.354  7.402  7.451  7.499 

48 

3-1 

7.548  7.596  7.645  7.694  7-744 

7.793  7.843  7.892  7.942  7.992 

49 

3-2 

8.042  8.093  8.143  8.194  8.245 

8.296  8.347  8.398  8.450  8.501 

5i 

3-3 

8.553  8.605  8.657  8.709  8.762 

8.814  8.867  8.920  8.973  9.026 

52 

3-4 

9.079  9,133  9.186  9.2.40  9.294 

9.348  9.402  9.457  9.511  9.566 

54 

3-5 

9.621  9.676  9.731  9.787  9.842 

9.898    9-954    IO.OI    IO.O7    IO.I2 

56 

3-6 

10.18  10.24  10.29  10.35  10.41 

10.46    10.52    10.58    10.64    10.69 

6 

3-7 

10.75  10-81  10.87  10.93  10.99 

II.O4    II-  IO    II.  l6    11.22    11.28 

6 

3-3 

11.34  H-4O  11.46  11.52  11.58 

11.64  11.70  11.76  11.82  11.88 

6 

3-9 

11.95  12.  or  12.07  12.13  12.19 

12.25  12-32  12.38  12.44  12.50 

6 

4.0 

12.57  !2.63  12.69  12.76  12.82 

12.88  12.95  13.01  13.07  13.14 

7 

4.1 

13.20  13.27  13.33  13-40  13-46 

13-53  13-59  13-66  13.72  13-79 

7 

4.2 

13.85  13.92  13.99  14-05  14.12 

14.19  14.25  14.32  14.39  14-45 

7 

4-3 

14.52  14.59  14-66  14.73  14.79 

14.86  14.93  15.00  15.07  15.14 

7 

4-4 

15.21  15.27  15.34  15.41  15.43 

15.55  15.62  15.69  15.76  15.83 

7 

4-5 

15.90  15.98  16.05  16.12  16.19 

16.26  16.33  16.40  16.47  16.55 

7 

4-6 

16.62  16.69  16.76  16.84  16.91 

16.98  17.06  17.13  17.20  17.28 

7 

4-7 

17.35  17.42  17.50  17.57  17.65 

17.72  17.80  17.87  17.95  18.02 

8 

4.8 

18.10  18.17  18.25  18.32  18.40 

18.47  18.55  18.63  18.70  18.78 

8 

4-9 

18.86  18.93  19.01   19.09  19.17 

19.24  19.32  19.40  19.48  19.56 

8 

5-0 

19.63  19.71  19.79  19.87  19.95 

2O.O3    2O.II    2O.I9    2O.27    2O.35 

8 

5-1 

20.43  20.51  20.59  20.67  20.75 

20.83  20.91  20.99  21.07  21.  16 

8 

5-2 

21.24  21.32  21.40  21.48  21.57 

21.65    21-73    2I.8I    2I.9O   21.98 

8 

5-3 

22.06  22.15  22.23  22.31  22.40 

22.48    22.56   22.65    22.73    22.82 

8 

5-4 

22.90  22.99  23.07  23.16  23.24 

23.33    23-41    23.50    23.59    23.67 

9 

d 

01234 

5          6          7          89 

Diff. 

AREAS   OF   CIRCLES. 


36i 


TABLE   VII.       AREAS    OF    CIRCLES. 


d 

01234 

56789 

Diff. 

5-5 

23.76  23.84  23.93  24.02  24.11 

24.19  24.28  24.37  24.45  24.54 

9 

5-6 

24.63  24.72  24.81  24.89  24.98 

25.07  25.16  25.25  25.34  25.43 

9 

5-7 

25.52  25.61  25.70  25.79  25-88 

25.97  26.06  26.15  26.24  26.33 

9 

5-8 

26.42  26.51  26.60  26.69  26.79 

26.88  26.97  27.06  27.15  27.25 

9 

5.9 

27.34  27.43  27.53  27.62  27.71 

27.81  27.90  27.99  28.09  27.18 

9 

6.0 

28.27  28.37  28.46  28.56  28.65 

28.75  28.84  28.94  29.03  29.13 

9 

6.1 

29.22  29.32  29.42  29.51  29.61 

29.71  29.80  29.90  30.00  30.09 

10 

6.2 

30.19  30.29  30.39  30.48  30.58 

30.68  30.78  30.^8  30.97  31.07 

10 

6-3 

31.17  31.27  31.37  31.47  31.57 

31.67  31.77  31.87  31.97  32.07 

10 

6.4 

32.17  32.27  32.37  32.47  32.57 

32.67  32.78  32.88  32.98  33-08 

10 

6.5 

33.18  33.29  33.39  33.49  33.59 

33.70  33.80  33.90  34  oo  34.ii 

10 

6.6 

34.21  34.32  34.42  34-52  34-63 

34-73  34.84  34-94  35.05  35.15 

10 

6.7 

35.26  35.36  35.47  35.57  35.68 

35.78  35.89  36.00  36.10  36.21 

10 

6.8 

36.32  36.42  36.53  36.64  36.75 

36.85  36.96  37.07  37.18  37-23 

ii 

6.9 

37-39  37.50  37-6r  37.72  37.83 

37.94  38.05  38.16  38.26  38.37 

ii 

7.0 

38.48  3S.59  3?-  70  38.82  38.93 

39.04  39.15  39.26  39.37  39.48 

ii 

7-1 

39-59  39-70  39.82  39.93  40.04 

40.15  40.26  40.38  40.49  40.60 

ii 

7.2 

40.72  40.83  40.94  41.06  41.17 

41.28  41.40  41.51  41.62  41-74 

ii 

7-3 

41.85  41.97  42.08  42.20  42.31 

42.43  42.54  42.66  42.78  42.89 

ii 

7-4 

43.01  43.12  43.24  43.36  43.47 

43-59  43.71  43.83  43-94  44-o6 

12 

7-5 

44.18  44-30  44-41  44-53  44.65 

44.77  44.89  45.01  45.13  45.25 

12 

7-6 

45.36  45-48  45.60  45.72  45.84 

43.96  46.08  46.20  46.32  46.45 

12 

7-7 

46.57  46.69  46.81  46.93  47.05 

47-17  47-29  47.42  47.54  47-66 

12 

7-8 

47.78  47.91  48.03  48.15  48.27 

48.40  48  52  48.65  48.77  48.89 

12 

7-9 

49.02  49.14  49.27  49.39  49.51 

49.64  49.76  49.89  50.01  50.14 

12 

8.0 

50.27  50.39  50.52  50.64  50.77 

50.90  51.02  51.15  51.28  51.40 

13 

8.1 

51.53  51.66  51.78  51.91  52.04 

52.17  52.30  52.42  52.55   52.68 

13 

8.2 

52.81  52.94  53.07  53.20  53.33 

53.46  53.59  53.72  53.85   53.98 

13 

8-3 

54.11  54.24  54.37  54.50  54.63 

54.76  54.89  55.02  55.15   55-29 

13 

8.4 

55.42  55.55  55.68  55.81  55.95 

56.08  56.21   56.35   56.48   56.61 

13 

8.5 

56.75  56.88  57.01  57.15  57.28 

57-41   57.55  57.68  57.82  57.95 

13 

8.6 

58.09  58.22  58.36  58.49  58.63 

58.77  58.90  59.04  59.17  59.31 

14 

8.7 

59-45  59.58  59-72  59-86  59-99 

60.13  60.27  60.41  60.55  60.68 

M 

8.8 

60.82  60.96  61.10  61.24  61.38 

61.51  61.65  61.79  61.93  62.07 

14 

8.9 

62.21  62.35  62.49  62.63  62.77 

62.91  63.05  63.19  63.33  63.48 

14 

9.0 

63.62  63.76  63.90  64.04  64.18 

64.33  64.47  64.61  64.75  64-90 

14 

9.1 

65.04  65.18  65.33  65.47  65.61 

65.76  65.90  66.04  66.19  66.33 

14 

9.2 

66.48  6662  66.77  66.91  67.06 

67.20  67.35  67.49  67.64  67.78 

15 

9-3 

67.93  68.08  68.22  68.37  68.51 

68.66  68.81  68.96  69.10  69.25 

15 

9.4 

69.40  69.55  69.69  69.84  69.99 

70.14  70.29  70.44  70.58  70.73 

15 

9-5 

70.88  71.03  71.18  71.33  71.48 

71.63  71.78  71.93  72.08  72.23 

15 

9.6 

72.38  72.53  72.68  72.84  72.99 

73  14  73.29  73.44  73.59  73.75 

15 

9-7 

73.90  74.05  74.20  74.36  74-51 

74.66  74.82  74.97  75.12  75.28 

15 

9.8 

75-43  75.58  75.74  75.89  76.05 

76.20  76.36  76.51  76.67  76.82 

16 

9-9 

76.98  77.13  77.29  77.44  77.60 

77-76  77.91  78.07  78.23  78.38 

16 

d 

G             I             2             3             4 

56789 

Diff. 

362 


APPENDIX. 


TABLE  VIII.— WEIGHT  OF  WROUGHT-IRON  BARS. 


Side  or 
Diam- 
eter. 
Inches. 

Pounds  per  Linear 
Foot. 

Side  or 
Diam- 
eter. 
Inches. 

Pounds  per  Linear 
Foot. 

Side  or 
Diam- 
eter. 
Inches. 

Pounds  per  Linear 
Foot. 

Square 
Bars. 

Round 
Bars. 

Square 
Bars. 

Round 
Bars. 

Square 
Bars. 

Round 
Bars. 

0 

2 

13.33 

10.47 

5 

83.33 

65.45 

ft 

0.013 

O.OIO 

ft 

14.18 

11.14 

i 

87.55 

68.76 

i 

0.052 

O.O4I 

1 

15,05 

11.82 

* 

91.88 

72.16 

ft 

O.II7 

0.092 

ft 

15.95 

12.531 

1 

96.30 

75.64 

i 

0.208 

0.164 

i 

16.88 

13.25 

i 

100.8 

79.19 

& 

0.326 

0.256 

ft 

17.83 

14.00 

f 

IQ5.5 

82.83 

1 

0.469 

0.368 

1 

18.80 

14-77 

i 

110.  2 

86.56 

ft 

0.638 

0.501 

TV 

19.80 

15.55 

1 

II5-I 

90.36 

* 

0.833 

0.654 

i 

20.83 

16.36 

6 

120.0 

94-25 

A 

1.055 

0.828 

& 

21.89 

17.19 

i 

I25.I 

98.22 

f 

1.302 

1.023 

f 

22.97 

18.04 

i 

130.2 

102.3 

tt 

1.576 

1.237 

H 

24.08 

18.91 

1 

135.5 

106.4 

f 

1.875 

1-473 

i 

25.21 

19.80 

1 

140.8 

no.  6 

it 

2.2OI 

1.728 

if 

26.37 

20.71 

f 

146.3 

114.9 

1 

2-552 

2.004 

1 

27-55 

21.64 

f 

I5I.9 

II9-3 

if 

2-930 

2.301 

it 

28.76 

22.59 

1 

157-6 

123-7 

i 

3-333 

2.618 

3 

30.00 

23.56 

7 

166.3 

128.3 

TV 

3.763 

2-955 

i 

32.55 

25.57 

i 

175-2 

137.6 

i 

4.219 

3-3I3 

i 

35-21 

27.65 

1 

187.5 

147-3 

ft 

4-701 

3.692 

f 

37-97 

29.82 

f 

200.2 

157-2 

i 

5.208 

4.091 

i 

40.83 

32.07 

8 

213.3 

167.6 

T5* 

5.742 

4.510 

f 

43.8o 

34.40 

i 

226.9 

178.2 

1 

6.302 

4.950 

f 

46.88 

36.82 

i 

240.8 

189.2 

T5* 

6.888 

5.410 

1 

50.05 

39-31 

f 

255.2 

200.4 

i 

7.500 

5.890 

4 

53-33 

41.89 

9 

270.0 

212.  1 

ft 

8.138 

6.392 

i 

56.72 

44-55 

i 

285.2 

224.0 

1 

8.802 

6.913 

i 

60.21 

47-29 

i 

300.8 

236.3 

« 

9.492 

7-455 

1 

63.80 

50.11 

f 

316.9 

248.9 

f 

10.21 

8.018 

i 

67.50 

53-01 

10 

333-3 

261.8 

it 

10.95 

8.601 

1 

71.30 

56.00 

i 

350.2 

275-1 

1 

11.72 

9.204 

f 

75-21 

59-07 

i 

367.5 

288.6 

it 

12,51 

9.828 

1 

79.22 

62.22 

f 

385.2 

302.5 

INDEX. 


363 


INDEX. 


Advanced  problems,  347 
Aluminum,  190 
Angle  iron,  127 
Angular  velocity,  233 
Annealing,  183,  187 
Answers  to  problems,  351 
Apparent  and  true  stresses,  288-309 

shears,  305 
Appendix,  344-362 
Areas  of  circles,  355,  360 
Army,  gun  formulas,  319,  320 
Average  constants,  163 

BACH'S  formulas,  332 
Bariron.  182,  362 
BARLOW'S  formula,  28 
Bars  of  uniform  strength,  227 
resilience  of,  202 
under  centrifugal  stress,  233 
under  impact,  229,  231 
weights  of,  2,  355,  362 
Beams,  36-110,   146-161,  243-271 

bending  moments,  42,  347 

cantilevers,  36-94 

cast  iron,  60 

center  of  gravity  of  sections,  52 

combined  stresses,  146-149,  267-271 

continuous,  99-110 

deck,  67 

definitions,  36 

deflection,  72-77,  95,245,  249,  265 

deflection  and  stiffness,  74 

deflection  and  stress,  79 

designing  of,  58 

elastic  curve,  70 

elastic  resilience,  203 

experimental  laws,  48 

fixed,  88-98 

flexure  of,  107,  243-271 

flexure  and  torsion,  152 

fundamental  formulas,  49 

GALILEI'S  investigations,  162 

horizontal  shear,  155 

impact  on,  249 

internal  stresses,  45,  158 

internal  work,  203 

maximum  moments,  52,  347 

modulus  of  rupture,  61 

moments  of  inertia,  53 


Beams,  moving  loads,  259,  348 

overhanging,  85,  91 

reactions,  37,  85 

resilience  of,  96,  243 

restrained,  85,  95 

safe  loads  for,  58 

simple,  36-84 

stiffness,  74 

sudden  loads,  248 

theoretical  laws,  48 

uniform  strength,  80-84 

vertical  shear,  39,  262 

weights  of,  2,  355,  362 
Bending  moment,  42-44 

maximum,  54 

maximum  maximorum,  348 

tables  of,  79,  95,  104 

triangular  load,  257 
Beton,  189 

BIRNIE'S  formulas,  319 
Boilers,  25,  31,  333 

ends  of,  328 

joints  in,  28-34 

tubes  in,  25 
Bolts,  17,  34,  281,  307 
Books  of  reference,  163,  166,  289,  326 
Brass,  190 
Brick,  172-174 

modulus  of  rupture,  62,  174 

strength  of,  14,  173 

weight  of,  2,  173,  355 
Brick  tower,  15 
Bridge  iron,  182 

rollers,  239 
Briquettes,  175 
Bronze,  190 
Butt  joints,  30,  31 

Cantilever  beams,  36-110 

deflection  of,  72,  82 

elastic  curve,  73 

fundamental  formulas,  49 

internal  work,  244 

resilience,  204 

tables  for,  79,  95 

uniform  ftrength,  80 
Carbon  in  steel,  180,  183,  186 
Castings.  179,  189 
Cast  iron,  constants  for,  163,  179 


INDEX. 


Cast  iron,  factors  of  safety,  18 

in  compression,  14,  180 
in  shear,  15 
in  tension,  9,  180 
modulus  of  rupture,  61, 181 
pipes,  22,  23 
resilience  of,  199,  219 
weight  of,  i,  355 
Cements,  174 
Center  of  gravity,  52 
Centrifugal  stress,  232,  255 
Chestnut,   171 

CHRISTIE'S  experiments,  129,  341 
Circles,  areas  of,  355,  360 
Circular  plates,  328 
CLAVARINO'S  formulas,  317 
Coefficient  of  elasticity  7,  8,  163 

compression,   14 

shear,  15,  275 

tension,  9,  163,  165 
Coefficient  of  expansion,  145 
Cold  bend  test,  182,  214,  224 

rolling,  183,  217 
Columns,  111-134,  337~343 

deflection  of,  133,  340 

design  of,  125,  337 

eccentric  loads,  133,  343 

ends  of,  113,  120 

EULER'S  formula,  114,  340 

experiments  on,  337 

GORDON'S  formula,  119 

HODGKINSON'S  formula,  117 

investigation  of,  123,  337 

JOHNSON'S  (T.  H.)  formula,  127 

modified  EULER  formula,  339 

radius  of  gyration,  122 

RANKINE'S  formula,  119 

RITTER'S  formula,  132 

rupture  of,  113,  127 

safe  loads  for,  124 

sections  of,  in 

theory  of,  131 
Combined  stresses,  144-161,  288-309 

compression  and  flexure,  148,  269 

flexure  and  torsion,  152 

shear  and  tension,  150,  296 

tension  and  compression,  144,  290 

tension  and  flexure,  146,  267 

torsion  and  compression,  154 
Comparison  of  beams,  62,  77,  94 

shafts,  279 

Compound  cylinder,  315,  323 
Compression,  3,  4,  13,  226 

and  flexure,  148,  269        ^ 

and  shear,  150 

and  tension,  144,  290 

and  torsion,  154 


Compression,  cast  iron,  180 

cement,  176 

eccentric  loads,  241,  342 

mortar,  176 

steel,  188 

stone,  177 
Concrete,  189 
Connecting  rod,  256 
Constants,  tables  of,  163 
Continuity,  96 
Continuous  beams,  36,  76-110 

equal  spans,  106 

properties  of,  99 

tables  of,  104,  105 

three  moments,  102 

unequal  spans,  106 
Contraction  of  area,  169,  215 
Couplings  for  shafts,  281 
Crank  arm,  283 

pin,  283,  285,  301 
Cross-sections,  52,  in,  124 
Cubic  equation,  299,  302 
Curve,  elastic,  71 

of  stresses,  10 
Cylinders,  22,  118,  310 

compound,  310,  315 

exterior  pressure,  24,  316 

interior  pressure,  22,  313 

thick,  26,  310 

thin,  24 

with  hoops,  235,  321 
Cylindrical  rollers,  238 

Deflection  of  beams,  36,  79,  95,  173 

cantilever  beams,  72-75 

restrained  beams,  85-95 

simple  beams,  75-99 

sudden  loads,  246 

under  impact,  251,  253 

under  moving  load,  259 

under  shearing,  264 
Deflection  of  columns,  115,  133,  340 
Deformations,  3-6,  290-292 
Description  of  tables,  355 
Designing,  12,  195 

beams,  58 

columns,  125,  337 

guns,  325 

shafts,  141,  280 
Detrusion,  3,  15,  262 
Ductile  materials,  162 
DUDLEY'S  tests,  219,  344 

Eccentric  loads,  134,  240,  342 
Efficiency  of  a  joint,  31 
Elastic  curve,  36,  165 
cantilever  beams,  73 


INDEX. 


Elastic  curve,  columns,  115,  343 

continuous  beams,  73 
general  equation,  70 
restrained  beams,  85-95 
simple  beams,  75 
Elastic  limit,  7,  165 

cast  iron,  180 

compression,  14 

shear,  15 

steel,  188 

tension,  9 

timber,  172 

wrought  iron,  183 
Elasticity,  laws  of,  6 

theory  of,  288 
Elastic  resilience,  197-225 
Ellipse  of  stress,  305 
Ellipsoid  of  stress,  297 
Elliptical  plates,  331 
Elongation,  3s  5,  8,  183 

ultimate,  9,  13,  168 

under  impact,  218,  229 

under  own  weight,  226 
ESTRADA'S  tests,  218,  223 
EULER'S  formula,  114,  340 
Exercises,  34,  109.  143,  347 
Experimental  laws,  6,  48,  191 
Experiments,  130,  134,  143,  164,  213 
External  work,  201,  243 

Factor  of  lateral  contraction,  276,  290, 

319 

of  safety,  17-21,  166,  213 
Fatigue  of  materials,  191 
Flexure,  36-110,  146-161,  243-271 

and  compression,  148,  269 

and  tension,  144,  267 

and  torsion,  152,  283 

erroneous  views,  109 

of  brick,  174 

of  cast  iron,  180 

of  crank  pin,  286 

under  impact,  210.  251 

under  live  load,  259 

work  of,  201,  243,  263 
Floor  beams,  66,  87 
Forge  pig,  179 
Formulas,  principal: 

(1)  P=AS, 5 

(2)  S  =  £s 8 

(3)  AS,  =  V, 50 


Formulas,  principal 

: 

(id)  P  —   Sc  a 

.  123 

*   '  A          7  ' 

1 

(.,)^=// 

C 

.  137 

(I2)*=-^L 

198  ooof 

,  •  •  • 

.  141 

(13),  151 

(M), 

157 

(15), 

184; 

(16),  186 

(17), 

194 

(18), 

200; 

(19),  204 

(20), 

231 

(21), 

240; 

(22),  250 

(23), 

259 

(24), 

264  ; 

(25),  276 

(26;, 

277 

(27), 

292  ; 

(28),  294 

(29), 

296 

(30), 

299; 

(31),  304 

(32), 

312 

(33), 

318; 

(34),  319 

(35), 

337- 

Foundry  pig,  179 

(4)7  = 


51 


(5)  ^.T=^7» 71 

(6),  (7),  (8),  (9), 97-103 


Glass,  191 

GORDON'S  formula,  119 

Granite,  177 

GRASHOF'S  formulas,  239,  330 

Gravity,  center  of,  52 

specific,  i,  180 
Gun  metal,  169 
Guns,  28,  310-327 

hooped,  315,  324 

solid,  313,  320 
Gyration,  radius  of,  124 

Helical  springs,  348 

Hemispheres,  333 

Hemlock,  171 

Historical  notes,  162,  208 

HODGKINSON'S  formula,  117 

Hollow  cylinders,  24,  26,  310-327 

shafts,  142,  278 

spheres,  24,  333 
HOOKE'S  law,  6,  164,  288 
Hoops,  centrifugal  stress,  234 

for  guns,  315,  321,  324 

shrinkage  of,  235,  316 
Horizontal  impact,  210,  229 

shear,  155,  273 
Horse-power,  140,  205 
Hydraulic  cement,  174 

I  beams,  51,  87,   106 
India  rubber,  289 
Inertia  of  a  bar,  220,  229 

of  a  beam,  250,  262 
moment  of,  53,  138 
Impact,  162-225 
on  bars,  210,  229 
on  beams,  211,  249 
pressure  due  to,  222,  253 


366 


INDEX. 


Impact,  tests  on,  215-220,  248,  344 
Impulse,  230 
Inflection  point,  85 
Internal  stresses,  4,  45,  158,  288 
work,  201,  209,  213,  244 
Investigation,  12,  14 

of  beams,  56,  259 

of  columns,  125,  337 

of  guns,  314,  323 

of  joints,  28 

of  shafts,  141,  284 
Iron,  168,  179,  181,  329 

Jacket  for  guns,  315 
Joints,  riveted,  28-34 
efficiency  of,  31 

KEEP'S  impact  machine,  219,  251 
KIRKALDY'S  tests,  215,  249 

LAME'S  formulas,  310,  335 
Lateral  contraction,  275,  290 

factor  of,  276 
Lateral  deflection,  134,  341 

deformation,  290 
LAUNHARDT'S  formula,  194 
Laws,  experimental,  6,  7,  48 
of  fatigue,  191 
of  internal  stresses,  46,  288 
of  resilience,  204,  208 
Lead,  190 
Limestone,  177 
Live  loads,  258 
Loads,  36,  240 

safe,  for  beams,  58 

safe,  for  columns,  126,  337 

sudden,  245 
Locomotive,  255 
X-ogarithms,  356 
Longitudinal  impact,  229 


Manholes,  331 

Materials,  constants  of,  162 

factors  of  safety,  18 
fatigue  of,  191 
resilience  of,  197-225 
strength  of,  162-196 
weights  of,  i,  355 
Maximum  internal  stresses,  158,  299, 

303,  306 

moments,  54,  347 
strength,  168 
Modulus  of  elasticity,  8 

resilience,  165,  199,  278 
rupture,  61,  174,  181,  185 
Moment  of  inertia,  53 
for  beams,  65,  67 


Moment  of  inertia,  for  columns,  116 

for  shafts,  138 
Moments,  42-107 

bending,  42 

for  continuous  beams,  104 

maximum,  54,  347 

resisting,  50,  136 

theorem  of  three,  102 

twisting,  137 
Mortar,  174 

Navy,  gun  formulas,  319,  326 
Neutral  axis,  49,  116 

surface,  48 
Nickel  steel,  188 

Oak,  171 

One-hoss  shay,  20 

Ordnance  formulas,  310-327 

Ores  of  iron,  179 

Oscillations  of  a  bar,  199,  210,  229 

of  a  beam,  246,  252 
Overhanging  beams,  85,  91 

Parabola,  43,  81,  133,  194 
Parallel  rod,  255 
Paving  brick,  173 
Phenomena  of  compression,  13 
shear,  15 
tension,  9 
torsion,  135 
Phosphor  bronze,  190 
Piers,  35,  228 
Pig  iron,  179 
Pine,  171 
Pipes,  22-27 
thick,  22 
thin,  26 
Piston  rod,  21 
Plate  girder,  259,  272,  308 

iron,  182 
Plates,  328-333 
on  cylinders,  238 
on  spheres,  237 
Polar  moments,  138 
Portland  cement,  175 
Power,  shafts  for,  140 
Pressure  due  to  impact   222,  253 
Principal  stresses,  299 
Problems,  2-350 
answers  to,  351 
Puddling  furnace,  181 

Radius  of  gyration,  122,  277 
Rafters,  148 

Railroad  rails,  2,  59,  219,  252,  254 
Range  of  stresses,  191-196 


INDEX. 


367 


Reactions,  37,  85,  101,  148 
Rectangular  beams,  63,  205 

plates,  332 

Repeated  stresses,  191-196,  217 
Resilience,  197-225 

of  bars,  202 

of  beams,  203,  243 

of  shearings,  274 

of  torsion,  277 

Resistance  of  materials,  1-22,  162-196 
Resisting  moment,  50,  136 
Restrained  beams,  85-95 

comparison  of,  94 
Resultant  stress,  295 
Riveted  joints,  28-35 

design  of,  32 

efficiency  of,  31 
Rivet  iron,  182 
Rivets,  28-34,  272 
Rollers,  236,  238 
Ropes,  189 

Round  shafts,  141,  279 
Rosendale  cement,  175 
Rupture,  4,  7,  9,  165 

of  beams,  51 

of  columns,  129 

in  repeated  stress,  191 

modulus  of,  61,  164 

Safe  loads,  58,  123,  337 

Safety,  factors  of,  17-21,  213 

Sandstone,  177 

Set,  6 

Shaft  couplings,  281 

Shafts,  135-143,  277-287 

for  power,  140 

hollow,  142,  278 

round,  141,  279 

square,  143 

resilience  of,  280 

stiffness  of,  279 

strength  of,  141,  278 
Shape  iron,  66,  182 
Shear,  3,  5,  15,  16,  172,  272-287 

and  tension,  16,  150 

horizontal,  155 

longitudinal,  15 

resilience  of,  274 

resisting,  47 

stresses  due  to,  272 

vertical,  39 
Shearing  stresses,  294,  303 

coefficient  of  elasticity,  266,  275 

deflection  due  to,  264 

ultimate  strength,  15,  308 

work  of,  263 
Shears  for  cantilevers,  42 


Shears  for  continuous  beams,  105 

for  simple  beams,  40 
Shocks,  7,  1 8,  199 
Short  beams,  265 
Shrinkage  of  hoops,  235,  321 
Simple  beams  ;  see  Beams. 
Slate,  178 

Solid  shafts,  141,  278 
Sound,  347 
Specific  gravities,  I 
Spheres,  24,  236,  333 
Spherical  rollers,  236 
Springs,  199.  348 
Square  plates,  333 
shafts,  143 

Squares  of  numbers,  355,  358 
Static  deflections,  246 

resilience,  208 
Statics,   288 
Steam  boilers,  24,  31,  333 

pipes,  22,  23 
Steel,  185-189 
constants  of,  163 
factors  of  safety,  18 
hoop  shrinkage,  235 
resilience,  200,  206 
Steel  beams,  66 
cranks,  284 
plates,  329 
rollers,  239 
spheres,  237     „ 
Stiffness  of  beams,  77 
of  shafts,  279 
Stone,  163,  177 
Straight- line  formula,  127 
Strain,  3 

Strength  of  materials,  1-21,  162-196 
history  of,  162 
tables,  164 
ultimate,  7,  168 
Stress,  ellipse  of,  305 

ellipsoid  of,  297 
velocity  of,  345 
Stresses,  3-21 
apparent,  288-309 
centrifugal,  232,  255 
combined,  144-161,  290,  308 
in  guns,  310-327 
repeated,  193 
sudden,  197,  247 
temperature,  145 
true,  288-309 
working,  17-21,  194 
Sudden  deflections,  246 
loads,  197 

Tables  ;  see  page  ix 


368 


INDEX. 


Temperature,  145,  235 
Tensile  tests,  n,  168 
Tension,  3,  4,  9,  226-242 
and  flexure,  146 
and  shear,  150 
apparent  and  true,  288-309 
cast  iron,  9,  180 
cement  and  mortar,  175 
eccentric,  240 
steel,  9,  187 
tangential,  312 
timber,  9,  171 
wrought  iron,  9,  183 
Testing  machines,  167-170,  213-220 

capacity  of,  170 
Tests,  ash  sticks,  109 
brick,  173,  174 
cast  iron,  180 
cement,  175 
cold  bend,  182 
columns,  127 
compression,  169 
continuous  beams,  no 
fatigue,  191 
impact,  211-224 
steel,  187 
stone,  178 
tension,  6,  n 
timber,  172 
torsion,  143,  170 
uniform  methods,  220 
wrought  iron,  n,  183 
Theorem  of  three  moments,  102 
Thick  hollow  cylinders,  26,  310 

spheres,  334 

THURSTON'S  torsion  machine,  143,  214 
Timber,  9,  14,  171 
factors  of  safety,  18 
modulus  of  rupture,  62 
resilience,  200 
weight,  i,  171,  355 
Torsion,  135-143,  272-287 
coefficient  of  elasticity,  139 
combined,  152,  154,  285 
modulus  of,  139 
phenomena  of,  135 
resilience  of,  277 

Transmission  of  power,  140,  285  t 
Trap  rock,  177 
True  deformations,  292 
stresses,  288-309 
Tubes  for  boilers,  25 
for  guns,  315 


Ultimate  resilience,  200,  206 
Ultimate  strength,  7,  168 

compression,  14,  169 

shear,  15 

tension,  14,  168 
Uniform  strength,  20 

bars,  227 

beams,  35,  80,  83 
Unit-stress,  3,  5 

repeated,  193 

working,  17 

Velocity  of  live  load,  257 

stress,  345 
Vertical  bar,  226 
Vertical  shear,  39-42,  105 
deflection  due  to,  264 
stresses  caused  by,  272 
work  of,  262 
Vibration  strength,  196 
Vibrations  of  a  beam,  252 
Volume,  change  of,  289 

resilience  of,  205,  212 

Water  pipes,  22,  23,  190 
Wave  propagation,  346 
Weights  of  bars,  2,  355,  362 
materials,  i,  163 
WEYRAUCH'S  formula,  194 
Wheel,  revolving,  234 
Wire,  183 
Wire  guns,  326 
W^HLER'S  tests,  191 
Work  of  flexure,  243 
rupture,  184 
shearing,  274 
tension,  202 
torsion,  278 
vertical  bar,  226 
vertical  shear,  262 
Working  stress,  12,  17,  192 
Wrought  iron,  163,  181 
factors  of  safety,  18 
resilience,  200 
shear,  15 
tension,  9,  183 
weight  of,  i,  362 
work  of,  184 
Wrought-iron  bars,  2,  36* 

Yield  point,  183 
YOUNG'S  laws,  208 


;  ,,«.  o5  CJ&***" 

r   -pfNE  O*  QET-UKN 

«v/eoDVJ£-  ^^===:=:::==:=^ 


THE  UNIVERSITY  OF  CALIFORNIA  UBRARV 


